6.3 Logarithmic Functions Do todos

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6.3 Logarithmic Functions Do todos

Learning Objectives

After completing this section, you should be able to…

Convert from logarithmic to exponential form.
Convert from exponential to logarithmic form.
Evaluate logarithms.
Use common logarithms.
Use natural logarithms.

Photo of the aftermath of the earthquake in Japan with a focus on the Japanese flag. — Figure 1 Devastation of March 11, 2011 earthquake in Honshu, Japan. (credit: Daniel Pierce)

In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes. One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings, like those shown in Figure 1. Even though both caused substantial damage, the earthquake in 2011 was 100 times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Haitian earthquake registered a 7.0 on the Richter Scale whereas the Japanese earthquake registered a 9.0.

The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 8 is not twice as great as an earthquake of magnitude 4. It is $10^{8-4} = 10^4 = 10,000$ times as great! In this lesson, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends.

Converting from Logarithmic to Exponential Form

In order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert between logarithmic and exponential form. For example, suppose the amount of energy released from one earthquake were 500 times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is $10^x = 500,$ where $x$ represents the difference in magnitudes on the Richter Scale. How would we solve for $x?$

We have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve $10^x = 500.$ We know that $10^2 = 100$ and $10^3 = 1000,$ so it is clear that $x$ must be some value between 2 and 3, since $y = 10^x$ is increasing. We can examine a graph, as in Figure 2, to better estimate the solution.

Graph of the intersections of the equations y=10^x and y=500. — Figure 2

Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph in Figure 2 passes the horizontal line test. The exponential function $y = b^x$ is one-to-one, so its inverse, $x = b^y$ is also a function. As is the case with all inverse functions, we simply interchange $x$ and $y$ and solve for $y$ to find the inverse function. To represent $y$ as a function of $x,$ we use a logarithmic function of the form $y = \log_{b}(x).$ The base $b$ logarithm of a number is the exponent by which we must raise $b$ to get that number.

We read a logarithmic expression as, “The logarithm with base $b$ of $x$ is equal to $y,$ ” or, simplified, “log base $b$ of $x$ is $y.$ ” We can also say, “ $b$ raised to the power of $y$ is $x,$ ” because logs are exponents. For example, the base 2 logarithm of 32 is 5, because 5 is the exponent we must apply to 2 to get 32. Since $2^5 = 32,$ we can write $\log_{2}32 = 5.$ We read this as “log base 2 of 32 is 5.”

We can express the relationship between logarithmic form and its corresponding exponential form as follows:

$\log_{b}(x) = y \Leftrightarrow b^y = x, b > 0, b \neq 1$

Note that the base $b$ is always positive.

Because logarithm is a function, it is most correctly written as $\log_{b}(x),$ using parentheses to denote function evaluation, just as we would with $f(x).$ However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses, as $\log_{b}x.$ Note that many calculators require parentheses around the $x.$

We can illustrate the notation of logarithms as follows:

Notice that, comparing the logarithm function and the exponential function, the input and the output are switched. This means $y = \log_{b}(x)$ and $y = b^x$ are inverse functions.

Definition of the Logarithmic Function

A logarithm base $b$ of a positive number $x$ satisfies the following definition.

For $x > 0, b > 0, b \neq 1,$

$y = \log_{b}(x) \text{ is equivalent to } b^y = x$ where,

we read $\log_{b}(x)$ as, “the logarithm with base $b$ of $x$ ” or the “log base $b$ of $x.$ “
the logarithm $y$ is the exponent to which $b$ must be raised to get $x.$

Also, since the logarithmic and exponential functions switch the $x$ and $y$ values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,

the domain of the logarithm function with base $b \text{ is } (0, \infty).$
the range of the logarithm function with base $b \text{ is } (-\infty, \infty).$

Q&A

Can we take the logarithm of a negative number?

No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.

How To

Given an equation in logarithmic form $\log_{b}(x) = y,$ convert it to exponential form.

Examine the equation $y = \log_{b}(x)$ and identify $b, y, \text{ and } x.$
Rewrite $\log_{b}(x) = y$ as $b^y = x.$

Example 1

Converting from Logarithmic Form to Exponential Form

Write the following logarithmic equations in exponential form.

$\log_{6}(\sqrt{6}) = \frac{1}{2}$
$\log_{3}(9) = 2$

Solution

First, identify the values of $b, y, \text{and} x.$ Then, write the equation in the form $b^y = x.$

$\log_{6}(\sqrt{6}) = \frac{1}{2}$ Here, $b = 6, y = \frac{1}{2}, \text{ and } x = \sqrt{6.}$ Therefore, the equation $\log_{6}(\sqrt{6}) = \frac{1}{2}$ is equivalent to $6^{\frac{1}{2}} = \sqrt{6}.$
$\log_{3}(9) = 2$ Here, $b = 3, y = 2, \text{ and } x = 9.$ Therefore, the equation $\log_{3}(9) = 2$ is equivalent to $3^2 = 9.$

Try It 1

Write the following logarithmic equations in exponential form.

$\log_{10}(1,000,000) = 6$
$\log_{5}(25) = 2$

Show Solution

$\log_{10}(1,000,000)=6$ is equivalent to $10^6=1,000,000$
$\log_5(25)=2$ is equivalent to $5^2=25$

Converting from Exponential to Logarithmic Form

To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base $b,$ exponent $x,$ and output $y.$ Then we write $x = \log_{b}(y).$

Example 2

Converting from Exponential Form to Logarithmic Form

Write the following exponential equations in logarithmic form.

$2^3 = 8$
$5^2 = 25$
$10^{-4} = \frac{1}{10,000}$

Solution

First, identify the values of $b, y, \text{and} x.$ Then, write the equation in the form $x = \log_{b}(y).$

$2^3 = 8$ Here, $b = 2,$ $x = 3,$ and $y = 8.$ Therefore, the equation $2^3 = 8$ is equivalent to $\log_{2}(8) = 3.$
$5^2 = 25$ Here, $b = 5,$ $x = 2,$ and $y = 25.$ Therefore, the equation $5^2 = 25$ is equivalent to $\log_{5}(25) = 2.$
$10^{-4} = \frac{1}{10,000}$ Here, $b = 10,$ $x = -4,$ and $y = \frac{1}{10,000}.$ Therefore, the equation $10^{-4} = \frac{1}{10,000}$ is equivalent to $\text{log}_{10}(\frac{1}{10,000}) = -4.$

Try It 2

Write the following exponential equations in logarithmic form.

$3^2 = 9$
$5^3 = 125$
$2^{-1} = \frac{1}{2}$

Show Solution

$3^2=9$ is equivalent to $\log_3(9)=2$
$5^3=125$ is equivalent to $\log_5(125)=3$
$2^{-1}=\frac{1}{2}$ is equivalent to $\log_2(\frac{1}{2})=-1$

Evaluating Logarithms

Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider $\log_{2}8.$ We ask, “To what exponent must $2$ be raised in order to get 8?” Because we already know $2^3 = 8,$ it follows that $\log_{2}8 = 3.$

Now consider solving $\log_{7}49$ and $\log_{3}27$ mentally.

We ask, “To what exponent must 7 be raised in order to get 49?” We know $7^2 = 49.$ Therefore, $\log_{7}49 = 2$
We ask, “To what exponent must 3 be raised in order to get 27?” We know $3^3 = 27.$ Therefore, $\log_{3}27 = 3$

Even some seemingly more complicated logarithms can be evaluated without a calculator. For example, let’s evaluate $\log_{\frac{2}{3}}\frac{4}{9}$ mentally.

We ask, “To what exponent must $\frac{2}{3}$ be raised in order to get $\frac{4}{9}?$ ” We know $2^2 = 4$ and $3^2 = 9,$ so $(\frac{2}{3})^2 = \frac{4}{9}.$ Therefore, $\log_{\frac{2}{3}}(\frac{4}{9}) = 2.$

How To

Given a logarithm of the form $y = \log_{b}(x),$ evaluate it mentally.

Rewrite the argument $x$ as a power of $b:$ $b^y = x.$
Use previous knowledge of powers of $b$ identify $y$ by asking, “To what exponent should $b$ be raised in order to get $x?$ ”

Example 3

Solving Logarithms Mentally

Solve $y = \log_{4}(64)$ without using a calculator.

Solution

First we rewrite the logarithm in exponential form: $4^y = 64.$ Next, we ask, “To what exponent must 4 be raised in order to get 64?”

We know $4^3 = 64$

Therefore, $\log_{4}(64) = 3$

Try It 3

Solve $y = \log_{121}(11)$ without using a calculator.

Show Solution

$\log_{121}(11)=\frac{1}{2}$ (recalling that $\sqrt{121}=(121)^{\frac{1}{2}}=11$ )

Example 4

Evaluating the Logarithm of a Reciprocal

Evaluate $y = \log_{3}(\frac{1}{27})$ without using a calculator.

Solution

First we rewrite the logarithm in exponential form: $3^y = \frac{1}{27}.$ Next, we ask, “To what exponent must 3 be raised in order to get $\frac{1}{27}?$ ”

We know $3^3 = 27,$ but what must we do to get the reciprocal, $\frac{1}{27}?$ Recall from working with exponents that $b^{-a} = \frac{1}{b^a}.$ We use this information to write

$\begin{array}{l} 3^{-3} = \frac{1}{3^3} \\ = \frac{1}{27} \end{array}$

Therefore, $\log_{3}(\frac{1}{27}) = -3.$

Try It 4

Evaluate $y = \log_{2}(\frac{1}{32})$ without using a calculator.

Show Solution

$\log_2(\frac{1}{32})=-5$

Using Common Logarithms

Sometimes you may see a logarithm written without a base. When you see one written this way, you need to look at the expression before evaluating it. It may be that the base you use doesn’t matter. If you find it in computer science, it often means $\log_{2}(x)$ . However, in mathematics it almost always means the common logarithm of 10. In other words, the expression $\log(x)$ often means $\log_{10}(x).$

Definition of the Common Logarithm

A common logarithm is a logarithm with base $10.$ We can also write $\log_{10}(x)$ simply as $\log(x).$ The common logarithm of a positive number $x$ satisfies the following definition.

For $x > 0,$

$y = \log(x) \text{ is equivalent to } 10^y = x$

We read $\log(x)$ as, “the logarithm with base $10$ of $x$ ” or “log base 10 of $x.$ ”

The logarithm $y$ is the exponent to which $10$ must be raised to get $x.$

Currently, we use $\log_{b}(x), \lg(x)$ as the common logarithm, $\text{lb}(x)$ as the binary logarithm, and $\ln(x)$ as the natural logarithm. Writing $\lg(x)$ without specifying a base is now considered bad form, despite being frequently found in older materials.

How To

Given a common logarithm of the form $y = \log(x),$ evaluate it mentally.

Rewrite the argument $x$ as a power of $10:$ $10^y = x.$
Use previous knowledge of powers of $10$ to identify $y$ by asking, “To what exponent must $10$ be raised in order to get $x?$ ”

Example 5

Finding the Value of a Common Logarithm Mentally

Evaluate $y = \log(1000)$ without using a calculator.

Solution

First we rewrite the logarithm in exponential form: $10^y = 1000.$ Next, we ask, “To what exponent must $10$ be raised in order to get 1000?” We know

$10^3 = 1000$

Therefore, $\log(1000) = 3.$

Try It 5

Evaluate $y = \log(1,000,000).$

Show Solution

$\log(1,000,000)=6$

How To

Given a common logarithm with the form $y = \log(x),$ evaluate it using a calculator.

Press [LOG].
Enter the value given for $x,$ followed by [ ) ].
Press [ENTER].

Example 6

Finding the Value of a Common Logarithm Using a Calculator

Evaluate $y = \log(321)$ to four decimal places using a calculator.

Solution

Press [LOG].
Enter 321, followed by [ ) ].
Press [ENTER].

Rounding to four decimal places, $\log(321) \approx 2.5065.$

Analysis

Note that $10^2 = 100$ and that $10^3 = 1000.$ Since 321 is between 100 and 1000, we know that $\log(321)$ must be between $\log(100)$ and $\log(1000).$ This gives us the following:

$\begin{array}{ccccc} 100 & < & 321 & < & 1000 \\ 2 & < & 2.5065 & < & 3 \end{array}$

Try It 6

Evaluate $y = \log(123)$ to four decimal places using a calculator.

Show Solution

$\log(123) \approx 2.0899$

Example 7

Rewriting and Solving a Real-World Exponential Model

The amount of energy released from one earthquake was 500 times greater than the amount of energy released from another. The equation $10^x = 500$ represents this situation, where $x$ is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?

Solution

We begin by rewriting the exponential equation in logarithmic form.

$\begin{array}{lll} 10^x & = 500 & \\ \log(500) & = x & \text{Use the definition of the common log.} \end{array}$

Next we evaluate the logarithm using a calculator:

Press [LOG].
Enter $500,$ followed by [ ) ].
Press [ENTER].
To the nearest thousandth, $\log(500) \approx 2.699.$

The difference in magnitudes was about $2.699.$

Try It 7

The amount of energy released from one earthquake was $\text{8,500}$ times greater than the amount of energy released from another. The equation $10^x = 8500$ represents this situation, where $x$ is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?

Show Solution

The difference in magnitudes was about 3.929.

Using Natural Logarithms

The most frequently used base for logarithms is $e,$ the value of which is approximately $2.71828$ . Base $e$ logarithms are important in calculus and some scientific applications; they are called natural logarithms. The base $e$ logarithm, $\log_{e}(x),$ has its own notation, $\ln(x).$

Most values of $\ln(x)$ can be found only using a calculator. The major exception is that, because the logarithm of 1 is always 0 in any base, $\ln 1 = 0.$ For other natural logarithms, we can use the $\ln$ key that can be found on most scientific calculators. We can also find the natural logarithm of any power of $e$ using the inverse property of logarithms.

Definition of the Natural Logarithm

A natural logarithm is a logarithm with base $e.$ We write $\log_{e}(x)$ simply as $\ln(x).$ The natural logarithm of a positive number $x$ satisfies the following definition.

For $x > 0,$

$y = \ln(x) \text{ is equivalent to } e^y = x$

We read $\ln(x)$ as, “the logarithm with base $e$ of $x$ ” or “the natural logarithm of $x.$ ”

The logarithm $y$ is the exponent to which $e$ must be raised to get $x.$

Since the functions $y = e^x$ and $y = \ln(x)$ are inverse functions, $\ln(e^x) = x$ for all $x$ and $e^{\ln(x)} = x$ for $x > 0.$

How To

Given a natural logarithm with the form $y = \ln(x),$ evaluate it using a calculator.

Press [LN].
Enter the value given for $x,$ followed by [ ) ].
Press [ENTER].

Example 8

Evaluating a Natural Logarithm Using a Calculator

Evaluate $y = \ln(500)$ to four decimal places using a calculator.

Solution

Press [LN].
Enter $500,$ followed by [ ) ].
Press [ENTER].

Rounding to four decimal places, $\ln(500) \approx 6.2146$

Try It 8

Evaluate $\ln(-500).$

Show Solution

It is not possible to take the logarithm of a negative number in the set of real numbers.

6.3 Exercise Set

What is a base $b$ logarithm? Discuss the meaning by interpreting each part of the equivalent equations $b^y = x$ and $\log_{b}x = y$ for $b > 0, b \neq 1.$
How is the logarithmic function $f(x) = \log_{b}x$ related to the exponential function $g(x) = b^x?$ What is the result of composing these two functions?
How can the logarithmic equation $\log_{b}x = y$ be solved for $x$ using the properties of exponents?
Discuss the meaning of the common logarithm. What is its relationship to a logarithm with base $b,$ and how does the notation differ?
Discuss the meaning of the natural logarithm. What is its relationship to a logarithm with base $b,$ and how does the notation differ?
$\text{log}_{4}(q) = m$
$\text{log}_{a}(b) = c$
$\log_{16}(y) = x$
$\log_{x}(64) = y$
$\log_{y}(x) = -11$
$\log_{15}(a) = b$
$\log_{y}(137) = x$
$\log_{13}(142) = a$
$\text{log}(v) = t$
$\text{ln}(w) = n$
$4^x = y$
$c^d = k$
$m^{-7} = n$
$19^x = y$
$x^{-\frac{10}{13}} = y$
$n^4 = 103$
$(\frac{7}{5})^m = n$
$y^x = \frac{39}{100}$
$10^a = b$
$e^k = h$
$\text{log}_{3}(x) = 2$
$\text{log}_{2}(x) = -3$
$\text{log}_{5}(x) = 2$
$\log_{3}(x) = 3$
$\text{log}_{2}(x) = 6$
$\text{log}_{9}(x) = \frac{1}{2}$
$\text{log}_{18}(x) = 2$
$\log_{6}(x) = -3$
$\text{log}(x) = 3$
$\text{ln}(x) = 2$
$\text{log}(100^8)$
$10^{\text{log}(32)}$
$2\text{log}(.0001)$
$e^{\ln(1.06)}$
$\ln(e^{-5.03})$
$e^{\ln(10.125)} + 4$
$\text{log}_{3}(\frac{1}{27})$
$\text{log}_{6}(\sqrt{6})$
$\text{log}_{2}(\frac{1}{8}) + 4$
$6\text{log}_{8}(4)$
$\text{log}(10,000)$
$\text{log}(0.001)$
$\text{log}(1) + 7$
$2\text{log}(100^{-3})$
$\text{ln}(e^{\frac{1}{3}})$
$\text{ln}(1)$
$\text{ln}(e^{-0.225}) - 3$
$25\text{ln}(e^{\frac{2}{5}})$
$\text{log}(0.04)$
$\text{ln}(15)$
$\text{ln}(\frac{4}{5})$
$\text{log}(\sqrt{2})$
$\text{ln}(\sqrt{2})$
Is $x = 0$ in the domain of the function $f(x) = \log(x)?$ If so, what is the value of the function when $x = 0?$ Verify the result.
Is $f(x) = 0$ in the range of the function $f(x) = \log(x)?$ If so, for what value of $x?$ Verify the result.
Is there a number $x$ such that $\ln x = 2?$ If so, what is that number? Verify the result.
Is the following true: $\frac{\log_{3}(27)}{\log_{4}(\frac{1}{64})} = -1?$ Verify the result.
Is the following true: $\frac{\ln(e^{1.725})}{\ln(1)} = 1.725?$ Verify the result.
The exposure index $EI$ for a camera is a measurement of the amount of light that hits the image receptor. It is determined by the equation $EI = \log_{2}(\frac{f^2}{t}),$ where $f$ is the “f-stop” setting on the camera, and $t$ is the exposure time in seconds. Suppose the f-stop setting is $8$ and the desired exposure time is $2$ seconds. What will the resulting exposure index be?
Refer to the previous exercise. Suppose the light meter on a camera indicates an $EI$ of $-2,$ and the desired exposure time is 16 seconds. What should the f-stop setting be?
The intensity levels I of two earthquakes measured on a seismograph can be compared by the formula $\log\frac{I_1}{I_2} = M_1 - M_2$ where $M$ is the magnitude given by the Richter Scale. In August 2009, an earthquake of magnitude 6.1 hit Honshu, Japan. In March 2011, that same region experienced yet another, more devastating earthquake, this time with a magnitude of 9.0. How many times greater was the intensity of the 2011 earthquake? Round to the nearest whole number.