4.3: Logarithmic Functions
In this section, you will:
- Convert from logarithmic to exponential form.
- Convert from exponential to logarithmic form.
- Evaluate logarithms.
- Use common logarithms.
- Use natural logarithms.
Figure \( \PageIndex{ 1 } \) 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 \( \PageIndex{ 1 } \). 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 \( \PageIndex{ 2 } \), to better estimate the solution.
Figure \( \PageIndex{ 2 } \)
Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph in Figure \( \PageIndex{ 2 } \) 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:\[\log_b( x ) = y \iff b^y = x, b > 0, b \neq 1. \nonumber \]Note that the base \(b\) is always positive.
Because the 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.
A logarithm base \(b\) of a positive number \(x\) satisfies the following definition.
For \(x > 0, b > 0, b \neq 1\),\[y =\log_b( x ) \text{ is equivalent to } b^y = x, \nonumber \]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 \(b\) is \((0, \infty )\).
- the range of the logarithm function with base \(b\) is \((-\infty , \infty )\).
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.
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 \(b\), \(y\), and \(x\). Then, write the equation in the form \( b^y = x \).
- Here, \(b=6\), \(y = \frac{1}{2}\), and \(x=\sqrt{6}\). Therefore, the equation \(\log_6 ( \sqrt{ 6 } ) = \frac{1}{2}\) is equivalent to\[ 6^{1/2} = \sqrt{6}. \nonumber \]
- Here, \(b=3\), \(y=2\), and \(x=9\). Therefore, the equation \(\log_3(9)=2\) is equivalent to \(3^2 =9\).
Write the following logarithmic equations in exponential form.
- \(\log_{10}(1,000,000)=6\)
- \(\log_5(25)=2\)
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 \( x = \log_b(y) \).
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 \( b \), \( y \), and \( x \). Then, write the equation in the form \( x = \log_b(y) \).
- Here, \(b=2\), \(x=3\), and \(y=8\). Therefore, the equation \( 2^3 = 8 \) is equivalent to \( \log_2(8) = 3 \).
- Here, \(b=5\), \(x=2\), and \(y=25\). Therefore, the equation \( 5^2 = 25 \) is equivalent to \( \log_5(25) = 2 \).
- Here, \(b=10\), \(x=−4\), and \( y = \frac{1}{10,000} \). Therefore, the equation \( 10^{-4} = \frac{1}{10,000} \) is equivalent to \( \log_{10} \left( \frac{1}{10,000} \right) = -4 \).
Write the following exponential equations in logarithmic form.
- \( 3^2 = 9 \)
- \( 5^3 = 125 \)
- \( 2^{-1} = \frac{1}{2} \)
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 simplifying \( \log_7 49 \) 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 \( \log_{2/3} \frac{4}{9} \) 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 \( \left( \frac{2}{3} \right)^2 = \frac{4}{9} \). Therefore, \( \log_{2/3} \left( \frac{4}{9} \right) = 2\).
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\[4^3 = 64. \nonumber \]Therefore,\[ \log_4 ( 64 ) = 3. \nonumber \]
Solve \(y = \log_{121} (11)\) without using a calculator.
Evaluate \(y = \log_3 \left( \frac{1}{27} \right)\) 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\[3^{−3} = \dfrac{1}{3^3} = \dfrac{1}{27}. \nonumber \]Therefore, \( \log_3 \left( \frac{1}{27} \right) = −3\).
Evaluate \(y = \log_2 \left( \frac{1}{32} \right)\) without using a calculator.
Using Common Logarithms
Sometimes we may see a logarithm written without a base. In this case, we assume that the base is \( 10 \). In other words, the expression \( \log(x)\) means \( \log_{10}(x) \). We call a base-10 logarithm a common logarithm . Common logarithms are used to measure the Richter Scale mentioned at the beginning of the section. Scales for measuring the brightness of stars, decibel levels of sounds, and the pH of acids and bases also use common logarithms.
A common logarithm is a logarithm with base \(10\). We write \( \log_{10}(x) \) simply as \( \log(x) \). The common logarithm of a positive number \(x\) satisfies the following definition.
For \(x > 0\),\[y = \log(x) \text{ is equivalent to } 10^y = x. \nonumber \]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 \(x\).
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\[10^3 = 1000. \nonumber \]Therefore, \( \log(1000)=3\).
Evaluate \(y= \log(1,000,000)\).
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, \( \log(321) \approx 2.5065\).
We can perform a quick "logic check" on our answer to Example \( \PageIndex{ 6 } \). 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 \( \log(100) = 2\) and \(\log(1000) = 3\).
Evaluate \(y= \log(123)\) to four decimal places using a calculator.
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.\[ \begin{array}{rrclcl} & 10^x & = & 500 & & \\[6pt] \implies & \log(500) & = & x & \quad & \left( \text{definition of the common logarithm} \right) \\[6pt] \implies & x & \approx & 2.699 & \quad & \left( \text{using a calculator} \right) \\[6pt] \end{array} \nonumber \]The difference in Richter Scale magnitudes was about \(2.699\).
The amount of energy released from one earthquake was 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?
Using Natural Logarithms
The most frequently used base for logarithms is \(e\). 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 \(\mathrm{ln}\) 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.
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 \(x > 0\),\[ y = \ln(x) \text{ is equivalent to } e^y = x. \nonumber \]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 \(x > 0\).
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, \( \ln(500) \approx 6.2146\)
Evaluate \(\ln(−500)\).
Access this online resource for additional instruction and practice with logarithms.