# 5.2: Logarithmic Functions - Mathematics

A population of 50 flies is expected to double every week, leading to a function of the form (f(x)=50(2)^{x}), where x represents the number of weeks that have passed. When will this population reach 500? Trying to solve this problem leads to:

[500=50(2)^{x} onumber] Dividing both sides by 50 to isolate the exponential
[10=2^{x} onumber]

While we have set up exponential models and used them to make predictions, you may have noticed that solving exponential equations has not yet been mentioned. The reason is simple: none of the algebraic tools discussed so far are sufficient to solve exponential equations. Consider the equation (2^{x} =10) above. We know that (2^{3} =8) and (2^{4} =16), so it is clear that (x) must be some value between 3 and 4 since (g(x)=2^{x}) is increasing. We could use technology to create a table of values or graph to better estimate the solution.

From the graph, we could better estimate the solution to be around 3.3. This result is still fairly unsatisfactory, and since the exponential function is one-to-one, it would be great to have an inverse function. None of the functions we have already discussed would serve as an inverse function and so we must introduce a new function, named log as the inverse of an exponential function. Since exponential functions have different bases, we will define corresponding logarithms of different bases as well.

Definition: logarithm

The logarithm (base (b)) function, written (log _{b} left(x ight)), is the inverse of the exponential function (base (b)), (b^{x}).

Since the logarithm and exponential “undo” each other (in technical terms, they are inverses), it follows that:

properties of logs: inverse properties

[log _{b} left(b^{x} ight)=x]

[b^{log _{b} x} =x]

Since log is a function, it is most correctly written as (log _b}(c)), using parentheses to denote function evaluation, just as we would with (f(c)). However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written as (log _b c).

logarithm equivalent to an exponential

The statement (b^{a} =c) is equivalent to the statement (log _{b} (c)=a).

Alternatively, we could show this by starting with the exponential function(c=b^{a}), then taking the log base (b) of both sides, giving (log _{b} (c)=log _{b} b^{a}). Using the inverse property of logs, we see that (log _{b} (c)=a).

Since log is a function, it is most correctly written as (log _{b} (c)), using parentheses to denote function evaluation, just as we would with (f(c)). However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written as (log _{b} c).

Example (PageIndex{1})

Write these exponential equations as logarithmic equations:

1. (2^{3} =8)
2. (5^{2} =25)
3. (10^{-4} =dfrac{1}{10000})

Solution

a) (2^{3} =8) is equivalent to (log _{2} (8)=3)

b) (5^{2} =25) is equivalent to (log _{5} (25)=2)

c) (10^{-4} =dfrac{1}{10000}) is equivalent to (log _{10} left(dfrac{1}{10000} ight)=-4)

Example (PageIndex{2})

Write these logarithmic equations as exponential equations:

a) (log _{6} left(sqrt{6} ight)=dfrac{1}{2}) b) (log _{3} left(9 ight)=2)

Solution

a) (log _{6} left(sqrt{6} ight)=dfrac{1}{2}) is equivalent to (6^{1/2} =sqrt{6})

b) (log _{3} left(9 ight)=2) is equivalent to (3^{2} =9)

Exercise (PageIndex{1})

Write the exponential equation (4^{2} =16) as a logarithmic equation.

[log _{4} left(16 ight)=2=log _{4} 4^{2} =2log _{4} 4 onumber]

By establishing the relationship between exponential and logarithmic functions, we can now solve basic logarithmic and exponential equations by rewriting.

Example (PageIndex{3})

Solve (log _{4} left(x ight)=2) for (x).

Solution

By rewriting this expression as an exponential, (4^{2} =x), so (x = 16).

Example (PageIndex{4})

Solve (2^{x} =10) for (x).

Solution

By rewriting this expression as a logarithm, we get (x=log _{2} (10)).

While this does define a solution, and an exact solution at that, you may find it somewhat unsatisfying since it is difficult to compare this expression to the decimal estimate we made earlier. Also, giving an exact expression for a solution is not always useful – often we really need a decimal approximation to the solution. Luckily, this is a task calculators and computers are quite adept at. Unluckily for us, most calculators and computers will only evaluate logarithms of two bases. Happily, this ends up not being a problem, as we will see briefly.

Definition: common and natural logarithms

The common log is the logarithm with base 10, and is typically written (log (x)).

The natural log is the logarithm with base (e), and is typically written (ln (x)).

Example (PageIndex{5})

Evaluate (log (1000)) using the definition of the common log.

Values of the common log
numbernumber as exponentiallog(number)
1000(10^3)3
100(10^2)2
10(10^1)1
1(10^0)0
0.1(10^{-1})-1
0.01(10^{-2})-2
0.001(10^{-3})-3

Solution

To evaluate (log (1000)), we can let (x=log (1000)), then rewrite into exponential form using the common log base of 10:

(10^{x} =1000.)

From this, we might recognize that 1000 is the cube of 10, so (x) = 3.

We also can use the inverse property of logs to write (log _{10} left(10^{3} ight)=3).

Exercise (PageIndex{2})

Evaluate (log (1,000,000)).

[log left(1000000 ight)=log left(10^{6} ight)=6 onumber]

Example (PageIndex{6})

Evaluate (ln left(sqrt{e} ight)).

Solution

We can rewrite (ln left(sqrt{e} ight)) as (ln left(e^{1/2} ight)). Since ln is a log base (e), we can use the inverse property for logs: [ln left(e^{1/2} ight)=log _{e} left(e^{1/2} ight)=dfrac{1}{2} onumber].

Example (PageIndex{7})

Evaluate log(500) using your calculator or computer.

Solution

Using a computer, we can evaluate (log (500)approx 2.69897)

## Graphs of Logarithms

Recall that the exponential function (f(x) = 2^x) produces this table of values

 (x) (f(x)) -3 -2 -1 0 1 2 3 (frac{1}{8}) (frac{1}{4}) (frac{1}{2}) 1 2 4 8

Since the logarithmic function “undoes” the exponential, (g(x) = log _2 (x)) produces the table of values

 (x) (g(x)) (frac{1}{8}) (frac{1}{4}) (frac{1}{2}) 1 2 4 8 -3 -2 -1 0 1 2 3

In this second table, notice that

1. As the input increases, the output increases.
2. As input increases, the output increases more slowly.
3. Since the exponential function only outputs positive values, the logarithm can only accept positive values as inputs, so the domain of the log function is ((0,infty)).
4. Since the exponential function can accept all real numbers as inputs, the logarithm can output any real number, so the range is all real numbers or (( - infty ,infty )).

Sketching the graph, notice that as the input approaches zero from the right, the output of the function grows very large in the negative direction, indicating a vertical asymptote at (x=0).

In symbolic notation we write:

as (x o 0^+, f(x) o -infty), and as (x o infty , f(x) o infty).

Graphical Features of the Logarithm

Graphically, in the function (g(x) = log_b(x)):

The graph has a horizontal intercept at (1, 0);

The graph has a vertical asymptote at (x=0);

The graph is increasing and concave down;

The domain of the function is (x > 0), or ((0,infty));

The range of the function is all real numbers, or ((- infty ,infty)).

When sketching a general logarithm with base (b), it can be helpful to remember that the graph will pass through the points ((1, 0)) and ((b, 1)).

To get a feeling for how the base affects the shape of the graph, examine the graphs below.

Notice that the larger the base, the slower the graph grows. For example, the common log graph, while it grows without bound, it does so very slowly. For example, to reach an output of 8, the input must be 100,000,000.

Another important observation made was the domain of the logarithm. Like the reciprocal and square root functions, the logarithm has a restricted domain which must be considered when finding the domain of a composition involving a log.

Example (PageIndex{8})

Find the domain of the function (f(x) = log (5 - 2x)).

Solution

The logarithm is only defined with the input is positive, so this function will only be defined when (5 - 2x > 0). Solving this inequality,

[egin{align*} - 2x &> - 5 x &< frac{5}{2} end{align*} ]

The domain of this function is (x < frac{5}{2}), or in interval notation, (left( - infty ,frac{5}{2} ight)).

## Logarithm Properties

To utilize the common or natural logarithm functions to evaluate expressions like (log _{2} (10)), we need to establish some additional properties.

properties of logs: exponent property

[log _{b} left(A^{r} ight)=rlog _{b} left(A ight)]

To show why this is true, we offer a proof:

Since the logarithmic and exponential functions are inverses, (b^{log _{b} A} =A).

Raising both sides to the r power, we get (A^{r} =left(b^{log _{b} A} ight)^{r}).

Utilizing the exponential rule that states(left(x^{p} ight)^{q} =x^{pq}), (A^{r} =left(b^{log _{b} A} ight)^{r} =b^{rlog _{b} A})

Taking the log of both sides, (log _{b} left(A^{r} ight)=log _{b} left(b^{rlog _{b} A} ight))

Utilizing the inverse property on the right side yields the result: (log _{b} left(A^{r} ight)=rlog _{b} A)

Example (PageIndex{9})

Rewrite (log _{3} left(25 ight)) using the exponent property for logs.

Solution

Since 25 = 5({}^{2}),

[log _{3} left(25 ight)=log _{3} left(5^{2} ight)=2log _{3} left(5 ight) onumber]

Example (PageIndex{10})

Rewrite (4ln (x))using the exponent property for logs.

Solution

Using the property in reverse, (4ln (x)=ln left(x^{4} ight)).

Exercise (PageIndex{3})

Rewrite using the exponent property for logs: (ln left(dfrac{1}{x^{2} } ight)).

[3. ln left(dfrac{1}{x^{2} } ight)=ln left(x^{-2} ight)=-2ln (x) onumber ]

The exponent property allows us to find a method for changing the base of a logarithmic expression.

properties of logs: change of base

[log _{b} left(A ight)=dfrac{log _{c} (A)}{log _{c} (b)} onumber]

Proof

Let (log _{b} left(A ight)=x).

Rewriting as an exponential gives (b^{x} =A).

Taking the log base c of both sides of this equation gives

[log _{c} b^{x} =log _{c} A onumber]

Now utilizing the exponent property for logs on the left side,

[xlog _{c} b=log _{c} A onumber]

Dividing, we obtain (x=dfrac{log _{c} A}{log _{c} b}). Replacing our original expression for (x),

[log _{b} A=dfrac{log _{c} A}{log _{c} b} onumber]

With this change of base formula, we can finally find a good decimal approximation to our question from the beginning of the section.

Example (PageIndex{11})

Evaluate (log _{2} (10)) using the change of base formula.

Solution

According to the change of base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base (e):

[log _{2} 10=dfrac{log _{e} 10}{log _{e} 2} =dfrac{ln 10}{ln 2} onumber]

Using our calculators to evaluate this,

[dfrac{ln 10}{ln 2} approx dfrac{2.30259}{0.69315} approx 3.3219 onumber]

This finally allows us to answer our original question – the population of flies we discussed at the beginning of the section will take 3.32 weeks to grow to 500.

Example (PageIndex{12})

Evaluate (log _{5} (100)) using the change of base formula.

Solution

We can rewrite this expression using any other base. If our calculators are able to evaluate the common logarithm, we could rewrite using the common log, base 10.

[log _{5} (100)=dfrac{log _{10} 100}{log _{10} 5} approx dfrac{2}{0.69897} =2.861 onumber]

While we can solve the basic exponential equation (2^{x} =10) by rewriting in logarithmic form and then using the change of base formula to evaluate the logarithm, the proof of the change of base formula illuminates an alternative approach to solving exponential equations.

Solving Exponential Equations

1. Isolate the exponential expressions when possible
2. Take the logarithm of both sides
3. Utilize the exponent property for logarithms to pull the variable out of the exponent
4. Use algebra to solve for the variable.

Example (PageIndex{13})

Solve (2^{x} =10) for (x).

Solution

Using this alternative approach, rather than rewrite this exponential into logarithmic form, we will take the logarithm of both sides of the equation. Since we often wish to evaluate the result to a decimal answer, we will usually utilize either the common log or natural log. For this example, we’ll use the natural log:

[ln left(2^{x} ight)=ln (10) onumber] Utilizing the exponent property for logs,
[xln left(2 ight)=ln (10) onumber] Now dividing by ln(2),
[x=dfrac{ln (10)}{ln left(2 ight)} approx 3.3219 onumber]

Notice that this result matches the result we found using the change of base formula.

Example (PageIndex{14})

In the first section, we predicted the population (in billions) of India (t) years after 2008 by using the function (f(t)=1.14(1+0.0134)^{t}). If the population continues following this trend, when will the population reach 2 billion?

Solution

We need to solve for time (t) so that (f(t) = 2).

[2=1.14(1.0134)^{t} onumber] Divide by 1.14 to isolate the exponential expression
[dfrac{2}{1.14} =1.0134^{t} onumber] Take the logarithm of both sides of the equation
[ln left(dfrac{2}{1.14} ight)=ln left(1.0134^{t} ight) onumber] Apply the exponent property on the right side
[ln left(dfrac{2}{1.14} ight)=tln left(1.0134 ight) onumber] Divide both sides by ln(1.0134)
[t=dfrac{ln left(dfrac{2}{1.14} ight)}{ln left(1.0134 ight)} approx 42.23 ext{ years} onumber]

If this growth rate continues, the model predicts the population of India will reach 2 billion about 42 years after 2008, or approximately in the year 2050.

Exercise (PageIndex{4})

Solve (5(0.93)^{x} =10).

[5(0.93)^{x} =10(0.93)^{x} =2ln left(0.93^{x} ight)=ln left(2 ight)xln left(0.93 ight)=ln left(2 ight)dfrac{ln (2)}{ln (0.93)} approx -9.5513 onumber]

Example (PageIndex{14})

Solve (5(1.07)^{3t} =2)

Solution

To start, we want to isolate the exponential part of the expression, the ((1.07)^{3t}), so it is alone on one side of the equation. Then we can use the log to solve the equation. We can use any base log; this time we’ll use the common log.

[5(1.07)^{3t} =2 onumber] Divide both sides by 5 to isolate the exponential
[(1.07)^{3t} =dfrac{2}{5} onumber] Take the log of both sides.
[log left((1.07)^{3t} ight)=log left(dfrac{2}{5} ight) onumber] Use the exponent property for logs
[3tlog left(1.07 ight)=log left(dfrac{2}{5} ight) onumber] Divide by (3log left(1.07 ight)) on both sides
[dfrac{3tlog left(1.07 ight)}{3log left(1.07 ight)} =dfrac{log left(dfrac{2}{5} ight)}{3log left(1.07 ight)} onumber] Simplify and evaluate
[t=dfrac{log left(dfrac{2}{5} ight)}{3log left(1.07 ight)} approx -4.5143 onumber]

Note that when entering that expression on your calculator, be sure to put parentheses around the whole denominator to ensure the proper order of operations:

log(2/5)/(3*log(1.07))

Some situations cannot be addressed using the properties already discussed. For these, we need some additional properties.

properties of logs

Sum of Logs Property:

[log _{b} left(A ight)+log _{b} left(C ight)=log _{b} (AC)]

Difference of Logs Property:

[log _{b} left(A ight)-log _{b} left(C ight)=log _{b} left(dfrac{A}{C} ight) ]

It’s just as important to know what properties logarithms do not satisfy as to memorize the valid properties listed above. In particular, the logarithm is not a linear function, which means that it does not distribute:

[log A + B e log A + log B. label{distr1}]

With these properties, we can rewrite expressions involving multiple logs as a single log, or break an expression involving a single log into expressions involving multiple logs.

Example (PageIndex{15})

Write (log _{3} left(5 ight)+log _{3} left(8 ight)-log _{3} left(2 ight)) as a single logarithm.

Solution

Using the sum of logs property on the first two terms,

[log _{3} left(5 ight)+log _{3} left(8 ight)=log _{3} left(5cdot 8 ight)=log _{3} left(40 ight) onumber]

This reduces our original expression to

[log _{3} left(40 ight)-log _{3} left(2 ight) onumber]

Then using the difference of logs property,

[log _{3} left(40 ight)-log _{3} left(2 ight)=log _{3} left(dfrac{40}{2} ight)=log _{3} left(20 ight) onumber]

Example (PageIndex{16})

Evaluate (2log left(5 ight)+log left(4 ight)) without a calculator by first rewriting as a single logarithm.

Solution

On the first term, we can use the exponent property of logs to write

[2log left(5 ight)=log left(5^{2} ight)=log left(25 ight) onumber]

With the expression reduced to a sum of two logs, (log left(25 ight)+log left(4 ight)), we can utilize the sum of logs property

[log left(25 ight)+log left(4 ight)=log (4cdot 25)=log (100) onumber]

Since (100 = 10^2), we can evaluate this log without a calculator:

[log (100)=log left(10^{2} ight)=2 onumber]

Example (PageIndex{17})

Rewrite (ln left(dfrac{x^{4} y}{7} ight)) as a sum or difference of logs

Solution

First, noticing we have a quotient of two expressions, we can utilize the difference property of logs to write

[ln left(dfrac{x^{4} y}{7} ight)=ln left(x^{4} y ight)-ln (7) onumber]

Then seeing the product in the first term, we use the sum property

[ln left(x^{4} y ight)-ln (7)=ln left(x^{4} ight)+ln (y)-ln (7) onumber]

Finally, we could use the exponent property on the first term

[ln left(x^{4} ight)+ln (y)-ln (7)=4ln (x)+ln (y)-ln (7) onumber]

## Log properties in solving equations

The logarithm properties often arise when solving problems involving logarithms.

Example (PageIndex{18})

Solve (log (50x+25)-log (x)=2).

Solution

In order to rewrite in exponential form, we need a single logarithmic expression on the left side of the equation. Using the difference property of logs, we can rewrite the left side:

[log left(dfrac{50x+25}{x} ight)=2 onumber]

Rewriting in exponential form reduces this to an algebraic equation:

[dfrac{50x+25}{x} =10^{2} =100 onumber] Multiply both sides by (x)
[50x+25=100x onumber] Combine like terms
[25=50x onumber] Divide by 50
[x=dfrac{25}{50} =dfrac{1}{2} onumber]

Checking this answer in the original equation, we can verify there are no domain issues, and this answer is correct.

Exercise (PageIndex{6})

Solve (log (x^2 - 4) = 1 + log (x + 2)).

[12 onumber]

## Important Topics of this Section

• The Logarithmic function as the inverse of the exponential function
• Writing logarithmic & exponential expressions
• Properties of logs
• Inverse properties
• Exponential properties
• Change of base
• Sum and difference of logs properties
• Common log
• Natural log
• Graph of the logarithmic function (domain and range)
• Solving exponential equations
• Solving equations using log rules

## 5.2: Logarithmic Functions - Mathematics

The method of the previous section for deciding whether there is a local maximum or minimum at a critical value is not always convenient. We can instead use information about the derivative $f'(x)$ to decide since we have already had to compute the derivative to find the critical values, there is often relatively little extra work involved in this method.

How can the derivative tell us whether there is a maximum, minimum, or neither at a point? Suppose that $f'(a)=0$. If there is a local maximum when $x=a$, the function must be lower near $x=a$ than it is right at $x=a$. If the derivative exists near $x=a$, this means $f'(x)>0$ when $x$ is near $a$ and $x a$, because $f$ slopes down from the local maximum as we move to the right. Using the same reasoning, if there is a local minimum at $x=a$, the derivative of $f$ must be negative just to the left of $a$ and positive just to the right. If the derivative exists near $a$ but does not change from positive to negative or negative to positive, that is, it is positive on both sides or negative on both sides, then there is neither a maximum nor minimum when $x=a$. See the first graph in figure 5.1.1 and the graph in figure 5.1.2 for examples.

Example 5.2.1 Find all local maximum and minimum points for $f(x)=sin x+cos x$ using the first derivative test. The derivative is $f'(x)=cos x-sin x$ and from example 5.1.3 the critical values we need to consider are $pi/4$ and $5pi/4$.

The graphs of $sin x$ and $cos x$ are shown in figure 5.2.1. Just to the left of $pi/4$ the cosine is larger than the sine, so $f'(x)$ is positive just to the right the cosine is smaller than the sine, so $f'(x)$ is negative. This means there is a local maximum at $pi/4$. Just to the left of $5pi/4$ the cosine is smaller than the sine, and to the right the cosine is larger than the sine. This means that the derivative $f'(x)$ is negative to the left and positive to the right, so $f$ has a local minimum at $5pi/4$.

## 5.2: Logarithmic Functions - Mathematics

Linear functions. These are functions of the form:

where m and b are constants. A typical use for linear functions is converting from one quantity or set of units to another. Graphs of these functions are straight lines . m is the slope and b is the y intercept. If m is positive then the line rises to the right and if m is negative then the line falls to the right. Linear functions are described in detail here.

Quadratic functions. These are functions of the form:

where a , b and c are constants. Their graphs are called parabolas . This is the next simplest type of function after the linear function. Falling objects move along parabolic paths. If a is a positive number then the parabola opens upward and if a is a negative number then the parabola opens downward. Quadratic functions are described in detail here.

--> Power functions. These are functions of the form:

where a and b are constants. They get their name from the fact that the variable x is raised to some power. Many physical laws (e.g. the gravitational force as a function of distance between two objects, or the bending of a beam as a function of the load on it) are in the form of power functions. We will assume that a = 1 and look at several cases for b :

The power b is a positive integer. See the graph to the right. When x = 0 these functions are all zero. When x is big and positive they are all big and positive. When x is big and negative then the ones with even powers are big and positive while the ones with odd powers are big and negative.

The power b is a negative integer. See the graph to the right. When x = 0 these functions suffer a division by zero and therefore are all infinite. When x is big and positive they are small and positive. When x is big and negative then the ones with even powers are small and positive while the ones with odd powers are small and negative.

The power b is a fraction between 0 and 1. See the graph to the right. When x = 0 these functions are all zero. The curves are vertical at the origin and as x increases they increase but curve toward the x axis.

The power function is discussed in detail here.

Polynomial functions. These are functions of the form:

where a n , a n &minus1 , &hellip , a 2 , a 1 , a 0 are constants. Only whole number powers of x are allowed. The highest power of x that occurs is called the degree of the polynomial. The graph shows examples of degree 4 and degree 5 polynomials. The degree gives the maximum number of &ldquo ups and downs &rdquo that the polynomial can have and also the maximum number of crossings of the x axis that it can have.

Polynomials are useful for generating smooth curves in computer graphics applications and for approximating other types of functions. Polynomials are described in detail here.

Rational functions. These functions are the ratio of two polynomials. One field of study where they are important is in stability analysis of mechanical and electrical systems (which uses Laplace transforms).

When the polynomial in the denominator is zero then the rational function becomes infinite as indicated by a vertical dotted line (called an asymptote ) in its graph. For the example to the right this happens when x = &minus2 and when x = 7.

When x becomes very large the curve may level off. The curve to the right levels off at y = 5.

The graph to the right shows another example of a rational function. This one has a division by zero at x = 0. It doesn't level off but does approach the straight line y = x when x is large, as indicated by the dotted line (another asymptote).

Exponential functions. These are functions of the form:

where x is in an exponent (not in the base as was the case for power functions) and a and b are constants. (Note that only b is raised to the power x not a .) If the base b is greater than 1 then the result is exponential growth. Many physical quantities grow exponentially (e.g. animal populations and cash in an interest-bearing account).

If the base b is smaller than 1 then the result is exponential decay. Many quantities decay exponentially (e.g. the sunlight reaching a given depth of the ocean and the speed of an object slowing down due to friction).

Exponential functions are described in detail here.

Logarithmic functions. There are many equivalent ways to define logarithmic functions. We will define them to be of the form:

where x is in the natural logarithm and a and b are constants. They are only defined for positive x . For small x they are negative and for large x they are positive but stay small. Logarithmic functions accurately describe the response of the human ear to sounds of varying loudness and the response of the human eye to light of varying brightness. Logarithmic functions are described in detail here.

Sinusoidal functions. These are functions of the form:

where a , b and c are constants. Sinusoidal functions are useful for describing anything that has a wave shape with respect to position or time. Examples are waves on the water, the height of the tide during the course of the day and alternating current in electricity. Parameter a (called the amplitude) affects the height of the wave, b (the angular velocity) affects the width of the wave and c (the phase angle) shifts the wave left or right. Sinusoidal functions are described in detail here.

If you found this page in a web search you won&rsquot see the

## Graphs of Logarithms

Recall that the exponential function produces this table of values

Since the logarithmic function “undoes” the exponential, produces the table of values

In this second table, notice that

1. As the input increases, the output increases.
2. As input increases, the output increases more slowly.
3. Since the exponential function only outputs positive values, the logarithm can only accept positive values as inputs, so the domain of the log function is .
4. Since the exponential function can accept all real numbers as inputs, the logarithm can output any real number, so the range is all real numbers or .

Sketching the graph, notice that as the input approaches zero from the right, the output of the function grows very large in the negative direction, indicating a vertical asymptote at x = 0.

In symbolic notation, we write

as , and as

Graphical Features of the Logarithm

Graphically, in the function

The graph has a horizontal intercept at (1, 0)

The graph has a vertical asymptote at x = 0

The graph is increasing and concave down

The domain of the function is x > 0, or

The range of the function is all real numbers, or

When sketching a general logarithm with base b, it can be helpful to remember that the graph will pass through the points (1, 0) and (b, 1).

To get a feeling for how the base affects the shape of the graph, examine the graphs below.

Notice that the larger the base, the slower the graph grows. For example, the common log graph, while it grows without bound, it does so very slowly. For example, to reach an output of 8, the input must be 100,000,000.

Another important observation made was the domain of the logarithm. Like the reciprocal and square root functions, the logarithm has a restricted domain which must be considered when finding the domain of a composition involving a log.

Find the domain of the function

The logarithm is only defined with the input is positive, so this function will only be defined when .

% Solving this inequality,

The domain of this function is , or in interval notation, .

Logarithm Properties

To utilize the common or natural logarithm functions to evaluate expressions like , we need to establish some additional properties.

Properties of Logs: Exponent Property

To show why this is true, we offer a proof.

Since the logarithmic and exponential functions are inverses,

% .

So

Utilizing the exponential rule that states ,

So then

Again utilizing the inverse property on the right side yields the result

Rewrite using the exponent property for logs.

Since 25 = 5 2 ,

Rewrite using the exponent property for logs.

Using the property in reverse,

3. Rewrite using the exponent property for logs: .

The exponent property allows us to find a method for changing the base of a logarithmic expression.

Properties of Logs: Change of Base

Let . Rewriting as an exponential gives . Taking the log base c of both sides of this equation gives

Now utilizing the exponent property for logs on the left side,

Dividing, we obtain

or replacing our expression for x,

With this change of base formula, we can finally find a good decimal approximation to our question from the beginning of the section.

Evaluate using the change of base formula.

According to the change of base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base e:

Using our calculators to evaluate this,

This finally allows us to answer our original question – the population of flies we discussed at the beginning of the section will take 3.32 weeks to grow to 500

Evaluate using the change of base formula.

We can rewrite this expression using any other base. If our calculators are able to evaluate the common logarithm, we could rewrite using the common log, base 10.

While we were able to solve the basic exponential equation by rewriting in logarithmic form and then using the change of base formula to evaluate the logarithm, the proof of the change of base formula illuminates an alternative approach to solving exponential equations.

Solving exponential equations:

Isolate the exponential expressions when possible

Take the logarithm of both sides

Utilize the exponent property for logarithms to pull the variable out of the exponent

Use algebra to solve for the variable.

Solve for x.

Using this alternative approach, rather than rewrite this exponential into logarithmic form, we will take the logarithm of both sides of the equation. Since we often wish to evaluate the result to a decimal answer, we will usually utilize either the common log or natural log. For this example, we’ll use the natural log:

Utilizing the exponent property for logs,

Now dividing by ln(2),

Notice that this result matches the result we found using the change of base formula.

In the first section, we predicted the population (in billions) of India t years after 2008 by using the function . If the population continues following this trend, when will the population reach 2 billion?

We need to solve for the t so that f(t) = 2

Divide by 1.14 to isolate the exponential expression

Take the logarithm of both sides of the equation

Apply the exponent property on the right side

Divide both sides by ln(1.0134)

If this growth rate continues, the model predicts the population of India will reach 2 billion about 42 years after 2008, or approximately in the year 2050.

4. Solve .

Some situations cannot be addressed using the properties already discussed. For these, we need some additional properties.

Sum of Logs Property:

Difference of Logs Property:

It’s just as important to know what properties logarithms do not satisfy as to memorize the valid properties listed above. In particular, the logarithm is not a linear function, which means that it does not distribute: log(A + B) ≠ log(A) + log(B).

With these properties, we can rewrite expressions involving multiple logs as a single log, or break an expression involving a single log into expressions involving multiple logs.

Write as a single logarithm.

Using the sum of logs property on the first two terms,

This reduces our original expression to

Then using the difference of logs property,

Evaluate without a calculator by first rewriting as a single logarithm.

On the first term, we can use the exponent property of logs to write

With the expression reduced to a sum of two logs, , we can utilize the sum of logs property

Since 100 = 10 2 , we can evaluate this log without a calculator:

Rewrite as a sum or difference of logs

First, noticing we have a quotient of two expressions, we can utilize the difference property of logs to write

Then seeing the product in the first term, we use the sum property

Finally, we could use the exponent property on the first term

5. Without a calculator evaluate by first rewriting as a single logarithm :

## 5.2: Logarithmic Functions - Mathematics

We begin with the exponential function defined by f ( x ) = 2 x and note that it passes the horizontal line test.

Therefore it is one-to-one and has an inverse. Reflecting y = 2 x about the line y = x we can sketch the graph of its inverse. Recall that if ( x , y ) is a point on the graph of a function, then ( y , x ) will be a point on the graph of its inverse.

To find the inverse algebraically, begin by interchanging x and y and then try to solve for y.

f ( x ) = 2 x y = 2 x ⇒ x = 2 y

We quickly realize that there is no method for solving for y. This function seems to “transcend” algebra. Therefore, we define the inverse to be the base-2 logarithm, denoted log 2 x . The following are equivalent:

This gives us another transcendental function defined by f − 1 ( x ) = log 2 x , which is the inverse of the exponential function defined by f ( x ) = 2 x .

The domain consists of all positive real numbers ( 0 , ∞ ) and the range consists of all real numbers ( − ∞ , ∞ ) . John Napier is widely credited for inventing the term logarithm.

In general, given base b > 0 where b ≠ 1 , the logarithm base b The exponent to which the base b is raised in order to obtain a specific value. In other words, y = log b x is equivalent to b y = x . is defined as follows:

## Mathematics - Logarithmic Functions and Their Graphs

This course teaches you all the promised concepts really well provided that you do have the prerequisites and do the exercises therein. Please see the "Prerequisites" and "What You Will Learn" sections.

The course has five main sections:

Section 1 - Course introduction - General information about the course

Section 2 - Logarithmic Functions - logarithmic functions and their relationship with exponential functions.

Section 3 - Logarithmic Functions Exercises - about fifty exercises to master the concepts in Section 2.

Section 4 - Graphs of Logarithmic Functions - basics (review) of function transformations and how to use them in order to graph logarithmic functions.

Section 5 - Graphs of Logarithmic Functions Exercises - about fifty exercises in order to master the concepts leaned in section 4.

This course can be used in conjunction with our subsequent courses to serve as a foundation to learn calculus (road map explained in the course).

The prerequisites of this course are algebra, functions in general and Exponential Functions. We have courses on Functions and Exponential Functions as well.

Please use the course code "41" to refer to this course in your correspondence with us. Thank you!

## 5.2: Logarithmic Functions - Mathematics

#### Weekly Plan - Spring 2010

DATE ACTIVITY TITLE
Feb 1–5 Ch 1 - Functions: Characteristics & Properties
1.1 Functions
1.2 Exploring absolute value
1.3 Properties of graphs of functions
1.4 Sketching graphs of functions

Feb 8–12
1.5 Inverse relations
1.6 Piecewise functions
1.7 Exploring operations with functions
Chapter review Test

Feb 16–19 Ch 2 – Functions: Understanding rates of change
2.1 Determine average rate of change
2.2 Estimating instantaneous rates of change from table of values & equations
2.3 Exploring instantaneous rates of change using graphs

Feb 22–26
2.4 Using rates of change to create a graphical model
2.5 Solving problems involving rates of change

Mar 1–5
Chapter review Test
Ch 3 – Polynomial functions
3.1 Exploring polynomial functions
3.2 Characteristics of polynomial functions

Mar 8–12
3.3 Characteristics of polynomial functions in factored form
3.4 Transformations of cubic & quartic functions
Mid chapter review

Mar 22–26
3.5 Dividing polynomials
3.6 Factoring polynomials
3.7 Factoring a sum or difference of cubes
Chapter review Test

Mar 29–Apr 1 Ch 4 – Polynomial equations & inequalities
4.1 Solving polynomial equations
4.2 Solving linear inequalities
4.3 Solving polynomial inequalities
4.4 Rates of change in polynomial functions

Apr 13–16
Chapter review Test
Ch 5 – Rational functions, equations & inequalities
5.1 Graphs of reciprocal functions
5.2 Exploring quotients of polynomial functions
5.3 Graphs of rational functions of the form f(x)=(ax+b)/(cx+d)

Apr 19-23
5.4 Solving rational equations
5.5 Solving rational inequalities
5.6 Rates of change in rational functions
Chapter review Test

Apr 26-30 Ch 6 – Trigonometric functions
6.2 Radian measure & angles on the Cartesian plane
6.3 Exploring graphs of the primary trigonometric functions
6.4 Transformations of trigonometric functions

May 3–7
Mid chapter review
6.6 Modeling with trigonometric functions
6.7 Rates of change in trigonometric functions
Chapter review Test

May 10–14 Ch 7 – Trigonometric identities and equations
7.1 Exploring equivalent trigonometric functions
7.2 Compound angle formulas
7.3 Double angle formulas

May 17–21
7.4 Proving trigonometric identities
7.5 Solving linear trigonometric equations
Chapter review Test

May 25–28 Ch 8 – Exponential & logarithmic functions
8.1 Exploring the logarithmic function
8.2 Transformations of logarithmic functions
8.3 Evaluating logarithms
8.4 Laws of logarithms

May 31–Jun 4
8.5 Solving exponential equations
8.6 Solving logarithmic equations
8.7 Solving problems with exponential and logarithmic functions
8.8 Rates of change in exponential & logarithmic functions

Jun 7–11
Chapter review Test
Ch 9 – Combinations of functions
9.1 Exploring combinations of functions
9.2 Combining two functions: Sums & differences
9.3 Combining two functions: Products

Jun 14–18
9.4 Exploring Quotients of functions
9.5 Composition of functions
9.6 Techniques for solving equations & inequalities
9.7 Modeling with functions

1. (log_3(8)-log_3(4))
2. (log_<10>(24)-log_<10>(6))
3. (log_begin p^3 end - log_begin p^2 end )
4. (log_5(64)-log_5(4)-log_5(2))
5. (log_begin p end + log_begin p^3 end - log_begin p^2 end)
1. (log_3(8)-log_3(4) = log_3(2))
2. (log_<10>(24)-log_<10>(6) = log_<10>(4))
3. (log_begin p^3 end - log_begin p^2 end = log_begin p end)
4. (log_5(64)-log_5(4)-log_5(2) = log_5(8))
5. (log_begin p end + log_begin p^3 end - log_begin p^2 end = log_begin p^2 end)

Select the question number you'd like to see the working for:

## 5.2: Logarithmic Functions - Mathematics

As with any word problem, the trick is convert a narrative statement or question to a mathematical statement.

Before we start, let's talk about earthquakes and how we measure their intensity.

In 1935 Charles Richter defined the magnitude of an earthquake to be

The magnitude of a standard earthquake is

Richter studied many earthquakes that occurred between 1900 and 1950. The largest had magnitude of 8.9 on the Richter scale, and the smallest had magnitude 0. This corresponds to a ratio of intensities of 800,000,000, so the Richter scale provides more manageable numbers to work with.

Each number increase on the Richter scale indicates an intensity ten times stronger. For example, an earthquake of magnitude 6 is ten times stronger than an earthquake of magnitude 5. An earthquake of magnitude 7 is times strong than an earthquake of magnitude 5. An earthquake of magnitude 8 is times stronger than an earthquake of magnitude 5.

Example 1: Early in the century the earthquake in San Francisco registered 8.3 on the Richter scale. In the same year, another earthquake was recorded in South America that was four time stronger. What was the magnitude of the earthquake in South American?

Solution: Convert the first sentence to an equivalent mathematical sentence or equation.

Convert the second sentence to an equivalent mathematical sentence or equation.

Example 2: A recent earthquake in San Francisco measured 7.1 on the Richter scale. How many times more intense was the San Francisco earthquake described in Example 1?

Solution: The intensity (I) of each earthquake was different. Let I 1 represent the intensity the early earthquake and I 2 represent the latest earthquake.

If you would like to work another example, click on example.

If you would like to test your knowledge by working some problems, click on problem.

## Contents

The powers of two have been known since antiquity for instance, they appear in Euclid's Elements, Props. IX.32 (on the factorization of powers of two) and IX.36 (half of the Euclid–Euler theorem, on the structure of even perfect numbers). And the binary logarithm of a power of two is just its position in the ordered sequence of powers of two. On this basis, Michael Stifel has been credited with publishing the first known table of binary logarithms in 1544. His book Arithmetica Integra contains several tables that show the integers with their corresponding powers of two. Reversing the rows of these tables allow them to be interpreted as tables of binary logarithms. [1] [2]

Earlier than Stifel, the 8th century Jain mathematician Virasena is credited with a precursor to the binary logarithm. Virasena's concept of ardhacheda has been defined as the number of times a given number can be divided evenly by two. This definition gives rise to a function that coincides with the binary logarithm on the powers of two, [3] but it is different for other integers, giving the 2-adic order rather than the logarithm. [4]

The modern form of a binary logarithm, applying to any number (not just powers of two) was considered explicitly by Leonhard Euler in 1739. Euler established the application of binary logarithms to music theory, long before their applications in information theory and computer science became known. As part of his work in this area, Euler published a table of binary logarithms of the integers from 1 to 8, to seven decimal digits of accuracy. [5] [6]

The binary logarithm function may be defined as the inverse function to the power of two function, which is a strictly increasing function over the positive real numbers and therefore has a unique inverse. [7] Alternatively, it may be defined as ln n/ln 2 , where ln is the natural logarithm, defined in any of its standard ways. Using the complex logarithm in this definition allows the binary logarithm to be extended to the complex numbers. [8]

As with other logarithms, the binary logarithm obeys the following equations, which can be used to simplify formulas that combine binary logarithms with multiplication or exponentiation: [9]

In mathematics, the binary logarithm of a number n is often written as log2 n . [10] However, several other notations for this function have been used or proposed, especially in application areas.

Some authors write the binary logarithm as lg n , [11] [12] the notation listed in The Chicago Manual of Style. [13] Donald Knuth credits this notation to a suggestion of Edward Reingold, [14] but its use in both information theory and computer science dates to before Reingold was active. [15] [16] The binary logarithm has also been written as log n with a prior statement that the default base for the logarithm is 2 . [17] [18] [19] Another notation that is often used for the same function (especially in the German scientific literature) is ld n , [20] [21] [22] from Latin logarithmus dualis [20] or logarithmus dyadis. [20] The DIN 1302 [de] , ISO 31-11 and ISO 80000-2 standards recommend yet another notation, lb n . According to these standards, lg n should not be used for the binary logarithm, as it is instead reserved for the common logarithm log10 n . [23] [24] [25]

### Information theory Edit

The number of digits (bits) in the binary representation of a positive integer n is the integral part of 1 + log2 n , i.e. [12]

In information theory, the definition of the amount of self-information and information entropy is often expressed with the binary logarithm, corresponding to making the bit the fundamental unit of information. However, the natural logarithm and the nat are also used in alternative notations for these definitions. [26]

### Combinatorics Edit

Although the natural logarithm is more important than the binary logarithm in many areas of pure mathematics such as number theory and mathematical analysis, [27] the binary logarithm has several applications in combinatorics:

• Every binary tree with n leaves has height at least log2n , with equality when n is a power of two and the tree is a complete binary tree. [28] Relatedly, the Strahler number of a river system with n tributary streams is at most log2n + 1 . [29]
• Every family of sets with n different sets has at least log2n elements in its union, with equality when the family is a power set. [30]
• Every partial cube with n vertices has isometric dimension at least log2n , and has at most 1 / 2 n log2n edges, with equality when the partial cube is a hypercube graph. [31]
• According to Ramsey's theorem, every n -vertex undirected graph has either a clique or an independent set of size logarithmic in n . The precise size that can be guaranteed is not known, but the best bounds known on its size involve binary logarithms. In particular, all graphs have a clique or independent set of size at least
• 1 / 2 log2n (1 − o(1)) and almost all graphs do not have a clique or independent set of size larger than 2 log2n (1 + o(1)) . [32]
• From a mathematical analysis of the Gilbert–Shannon–Reeds model of random shuffles, one can show that the number of times one needs to shuffle an n -card deck of cards, using riffle shuffles, to get a distribution on permutations that is close to uniformly random, is approximately
• 3 / 2 log2n . This calculation forms the basis for a recommendation that 52-card decks should be shuffled seven times. [33]

### Computational complexity Edit

The binary logarithm also frequently appears in the analysis of algorithms, [19] not only because of the frequent use of binary number arithmetic in algorithms, but also because binary logarithms occur in the analysis of algorithms based on two-way branching. [14] If a problem initially has n choices for its solution, and each iteration of the algorithm reduces the number of choices by a factor of two, then the number of iterations needed to select a single choice is again the integral part of log2 n . This idea is used in the analysis of several algorithms and data structures. For example, in binary search, the size of the problem to be solved is halved with each iteration, and therefore roughly log2 n iterations are needed to obtain a solution for a problem of size n . [34] Similarly, a perfectly balanced binary search tree containing n elements has height log2(n + 1) − 1 . [35]

The running time of an algorithm is usually expressed in big O notation, which is used to simplify expressions by omitting their constant factors and lower-order terms. Because logarithms in different bases differ from each other only by a constant factor, algorithms that run in O(log2 n) time can also be said to run in, say, O(log13 n) time. The base of the logarithm in expressions such as O(log n) or O(n log n) is therefore not important and can be omitted. [11] [36] However, for logarithms that appear in the exponent of a time bound, the base of the logarithm cannot be omitted. For example, O(2 log2 n ) is not the same as O(2 ln n ) because the former is equal to O(n) and the latter to O(n 0.6931. ) .

Algorithms with running time O(n log n) are sometimes called linearithmic. [37] Some examples of algorithms with running time O(log n) or O(n log n) are:

Binary logarithms also occur in the exponents of the time bounds for some divide and conquer algorithms, such as the Karatsuba algorithm for multiplying n -bit numbers in time O(n log2 3 ) , [42] and the Strassen algorithm for multiplying n × n matrices in time O(n log2 7 ) . [43] The occurrence of binary logarithms in these running times can be explained by reference to the master theorem for divide-and-conquer recurrences.

### Bioinformatics Edit

In bioinformatics, microarrays are used to measure how strongly different genes are expressed in a sample of biological material. Different rates of expression of a gene are often compared by using the binary logarithm of the ratio of expression rates: the log ratio of two expression rates is defined as the binary logarithm of the ratio of the two rates. Binary logarithms allow for a convenient comparison of expression rates: a doubled expression rate can be described by a log ratio of 1 , a halved expression rate can be described by a log ratio of −1 , and an unchanged expression rate can be described by a log ratio of zero, for instance. [44]

Data points obtained in this way are often visualized as a scatterplot in which one or both of the coordinate axes are binary logarithms of intensity ratios, or in visualizations such as the MA plot and RA plot that rotate and scale these log ratio scatterplots. [45]

### Music theory Edit

In music theory, the interval or perceptual difference between two tones is determined by the ratio of their frequencies. Intervals coming from rational number ratios with small numerators and denominators are perceived as particularly euphonious. The simplest and most important of these intervals is the octave, a frequency ratio of 2:1 . The number of octaves by which two tones differ is the binary logarithm of their frequency ratio. [46]

To study tuning systems and other aspects of music theory that require finer distinctions between tones, it is helpful to have a measure of the size of an interval that is finer than an octave and is additive (as logarithms are) rather than multiplicative (as frequency ratios are). That is, if tones x , y , and z form a rising sequence of tones, then the measure of the interval from x to y plus the measure of the interval from y to z should equal the measure of the interval from x to z . Such a measure is given by the cent, which divides the octave into 1200 equal intervals ( 12 semitones of 100 cents each). Mathematically, given tones with frequencies f1 and f2 , the number of cents in the interval from f1 to f2 is [46]

The millioctave is defined in the same way, but with a multiplier of 1000 instead of 1200 . [47]

### Sports scheduling Edit

In competitive games and sports involving two players or teams in each game or match, the binary logarithm indicates the number of rounds necessary in a single-elimination tournament required to determine a winner. For example, a tournament of 4 players requires log2 4 = 2 rounds to determine the winner, a tournament of 32 teams requires log2 32 = 5 rounds, etc. In this case, for n players/teams where n is not a power of 2, log2 n is rounded up since it is necessary to have at least one round in which not all remaining competitors play. For example, log2 6 is approximately 2.585 , which rounds up to 3 , indicating that a tournament of 6 teams requires 3 rounds (either two teams sit out the first round, or one team sits out the second round). The same number of rounds is also necessary to determine a clear winner in a Swiss-system tournament. [48]

### Photography Edit

In photography, exposure values are measured in terms of the binary logarithm of the amount of light reaching the film or sensor, in accordance with the Weber–Fechner law describing a logarithmic response of the human visual system to light. A single stop of exposure is one unit on a base- 2 logarithmic scale. [49] [50] More precisely, the exposure value of a photograph is defined as

where N is the f-number measuring the aperture of the lens during the exposure, and t is the number of seconds of exposure. [51]

Binary logarithms (expressed as stops) are also used in densitometry, to express the dynamic range of light-sensitive materials or digital sensors. [52]