5.4: Logarithmic Functions
- Convert from logarithmic to exponential form.
- Convert from exponential to logarithmic form.
- Evaluate logarithms.
- Use common logarithms.
- Use natural logarithms.
- Finding inverses of exponential and logarithmic functions.
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.
Figure \(\PageIndex{1}\): Devastation of March 11, 2011 earthquake in Honshu, Japan. (credit: Daniel Pierce).
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
\[10^{8−4}=10^4=10,000 \nonumber\]
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 \({\log}_232=5\). 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:
\[\begin{align*} \log_b(x)=y\Leftrightarrow b^y=x, b> 0, b\neq 1 \end{align*}\nonumber \]
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 \(\log_bx\). 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 \(x>0\), \(b>0\), \(b≠1\),
\[\begin{align*} y={\log}_b(x)\text{ is equivalent to } b^y=x \end{align*} \nonumber \]
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 \((−\infty,\infty)\).
We call the input, \(x\), the argument of the function.
Write the following logarithmic equations in exponential form.
- \({\log}_3(9)=2\)
- \({\log}_6(\sqrt{6})=\dfrac{1}{2}\)
- \({\log}_5(1)=0\)
- \({\log}_{10}(1,000,000)=6\)
Solution
First, identify the values of \(b\), \(y\),and \(x\). Then, write the equation in the form \(b^y=x\).
-
\({\log}_3(9)=2\)
Here, \(b=3\), \(y=2\),and \(x=9\). Therefore, the equation \({\log}_3(9)=2\) is equivalent to \[3^2=9\nonumber\]
-
\({\log}_6(\sqrt{6})=\dfrac{1}{2}\)
Here, \(b=6\), \(y=\dfrac{1}{2}\),and \(x=\sqrt{6}\). Therefore, the equation \({\log}_6(\sqrt{6})=\dfrac{1}{2}\) is equivalent to \[6^{\tfrac{1}{2}}=\sqrt{6}\nonumber\]
-
\({\log}_5(1)=0\)
Here, \(b=5\), \(y=0\),and \(x=1\). Therefore, the equation \({\log}_5(1)=0\) is equivalent to
\(5^0=1\)
Write the following logarithmic equations in exponential form.
- \({\log}_{10}(1,000,000)=6\)
- \({\log}_5(25)=2\)
- Answer a
-
\({\log}_{10}(1,000,000)=6\) is equivalent to \({10}^6=1,000,000\)
- Answer b
-
\({\log}_5(25)=2\) is equivalent to \(5^2=25\)
By establishing the relationship between exponential and logarithmic functions, we can now solve basic logarithmic equations by rewriting them in exponential form.
Solve \(\log _{4} \left(x\right)=2\) for \(x\).
Solution
By rewriting this expression as an exponential, \(4^{2} =x\), so \(x = 16\).
Solve \(\log _{2} \left(t\right)=5\) for \(t\).
- Answer
-
\(t=32\)
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.
- \(2^3=8\)
- \(4^2=16\)
- \({10}^{−4}=\dfrac{1}{10,000}\)
Solution
First, identify the values of \(b\), \(y\),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 \({\log}_2(8)=3\).
-
\(4^2=16\)
Here, \(b=4\), \(x=2\),and \(y=16\). Therefore, the equation \(4^2=16\) is equivalent to \({\log}_4(16)=2\).
-
\({10}^{−4}=\dfrac{1}{10,000}\)
Here, \(b=10\), \(x=−4\),and \(y=\dfrac{1}{10,000}\). Therefore, the equation \({10}^{−4}=\dfrac{1}{10,000}\) is equivalent to \({\log}_{10} \left (\dfrac{1}{10,000} \right )=−4\).
Write the following exponential equations in logarithmic form.
- \(3^2=9\)
- \(5^3=125\)
- \(2^{−1}=\dfrac{1}{2}\)
- Answer a
-
\(3^2=9\) is equivalent to \({\log}_3(9)=2\)
- Answer b
-
\(5^3=125\) is equivalent to \({\log}_5(125)=3\)
- Answer c
-
\(2^{−1}=\dfrac{1}{2}\) is equivalent to \({\log}_2 \left (\dfrac{1}{2} \right )=−1\)
By establishing the relationship between exponential and logarithmic functions, we can now solve basic exponential equations also.
Solve \(2^{x} =10\) for \(x\).
Solution
By rewriting this expression as a logarithm, we get \(x=\log _{2} (10)\).
Solve \(3^{m} =11\) for \(m\).
- Answer
-
\(m=\log_{3}(11)\)
Evaluating Logarithms
Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider \({\log}_28\). We ask, “To what exponent must \(2\) be raised in order to get 8?” Because we already know \(2^3=8\), it follows that \({\log}_28=3\).
Now consider solving \({\log}_749\) and \({\log}_327\) mentally.
- We ask, “To what exponent must \(7\) be raised in order to get \(49\)?” We know \(7^2=49\). Therefore, \({\log}_749=2\)
- 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 \(\log_{\ce{2/3}} \frac{4}{9}\) mentally.
- We ask, “To what exponent must \(\ce{2/3}\) be raised in order to get \(\ce{4/9}\)? ” We know \(2^2=4\) and \(3^2=9\), so \[{\left(\dfrac{2}{3} \right )}^2=\dfrac{4}{9}. \nonumber\] Therefore, \[{\log}_{\ce{2/3}} \left (\dfrac{4}{9} \right )=2. \nonumber\]
Evaluate the following without using a calculator:
- \({\log}_4(64)\)
- \({\log}_3 \left (\dfrac{1}{27} \right )\)
- \({\log}_5(-3)\)
Solution
- First let's set the expression equal to \(y\), \(y={\log}_4(64)\), then we can 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
\(4^3=64\)
So
\(4^y=4^3\)
Therefore,
\({\log}_4(64)=3\)
-
First, set the expression equal to a variable, \(y={\log}_3 \left (\dfrac{1}{27} \right )\), Then rewrite the logarithm in exponential form: \(3^y=\dfrac{1}{27}\). Next, we ask, “To what exponent must \(3\) be raised in order to get \(\dfrac{1}{27}\)?”
We know \(3^3=27\),but what must we do to get the reciprocal, \(\dfrac{1}{27}\)? Recall from working with exponents that \(b^{−a}=\dfrac{1}{b^a}\). We use this information to write
\[\begin{align*} 3^{-3}&= \dfrac{1}{3^3}\\ &= \dfrac{1}{27} \end{align*}\nonumber\]
Therefore, \({\log}_3 \left (\dfrac{1}{27} \right )=−3\).
- Logarithms are only defined for positive values so \(y={\log}_5(-3)\) is undefined.
Evaluate the following without using a calculator:
- \({\log}_{121}(11)\)
- \({\log}_2 \left (\dfrac{1}{32} \right )\)
- Answer
-
- \({\log}_{121}(11)=\dfrac{1}{2}\)
- \({\log}_2 \left (\dfrac{1}{32} \right )=−5\)
Properties of Logarithms
We saw above that it was convenient to convert to exponential form to evaluate logarithms. Recall that \({\log}_4(64)=3\). We know that \(4^3=64\), so we substitute this in, we get \[{\log}_4(64)={\log}_4(4^3)=3\nonumber\]
This gives us our first property for logarithms:
\( \log_b(b^x) =x \)
Next, consider \(x={\log}_31\). Converting to exponential form, \(3^x=1\). From properties of exponents we know \(x=0\). We can conclude that \({\log}_31\=0). This bring sus to our second property for logarithms:
\(\log_b1 =0\)
Similarly, if we convert \(x={\log}_bb\) to exponetial form we will find that \(x={\log}_bb=1\). This is our third property for logarithms:
\({\log}_bb=1\)
To evaluate \(x=4^{\log}_4(9)\), we can rewrite in logarithmic form as \({\log}_49={\log}_4x\). This means \(x=9\). This give us the following property:
\(b^{\log_b x}=x\) where \(x>0\)
- \(log_b(b^x) =x\)
- \(\log_b1 =0\)
- \({\log}_bb=1\)
- \(b^{\log_b x}=x\) where \(x>0\)
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 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 \(x>0\),
\[\begin{align*} y={\log}(x)\text{ is equivalent to } {10}^y=x \end{align*}\nonumber\]
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
\({10}^3=1000\)
Therefore, \(\log(1000)=3\).
Evaluate \(y=\log(1,000,000)\) without using a calculator.
- Answer
-
\(\log(1,000,000)=6\)
Evaluate \(y=log(321)\) and round to four decimal places using a calculator.
Solution
Rounding to four decimal places, \(\log(321)≈2.5065\).
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:
\(100<321<1000\)
\(2<2.5065<3\)
When using a calculator when approximating logarithms, it is important to make sure you input everything properly including having parentheses in the corect place.
Use your calculator to approximate \(log3-4\) to four decimal places.
Solution
Remember that sometimes parentheses are not written, \(log3-4\) is equivalent to \(log(3)-4\).
\(log3-4 \approx -3.5229\)
Evaluate \(y=\log(123)\) and round to four decimal places using a calculator.
- Answer
-
\(\log(123)≈2.0899\)
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. Rounding to three decimal places, what was the difference in magnitudes?
Solution
We begin by rewriting the exponential equation in logarithmic form.
\({10}^x=500\)
\(\log(500)=x\) Use the definition of the common log.
Next we evaluate the logarithm using a calculator:
To the nearest thousandth, \(\log(500)≈2.699\).
The difference in 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?
- Answer
-
The difference in magnitudes was about \(3.929\).
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, \(\ln1=0\). 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.
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 \(x>0\),
\(y=\ln(x)\) 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 \(x>0\).
Approximate \(y=\ln(500)\) to four decimal places using a calculator.
Solution
Rounding to four decimal places, \(\ln(500)≈6.2146\)
Approximate \(\ln(−500)\) to four decimal places using a calculator.
- Answer
-
It is not possible to take the logarithm of a negative number so \(\ln(−500)\) is undefined.
Solve \(\ln \left(a\right)=7\) for \(a\).
- Answer
-
\(a=e^7\)
Note that this answer is exact as \(e\) is an irrational number. To represent as a decimal approximation when asked, use your calculator to approximate.
There a couple of things worth mentioning from what we have learned about logarithms in this section.
- Logarithms are functions, \(f(x)={\log}_b(x)\). We call the input, \(x\), the argument of the function.
- When evaluating, sometimes we do not write parentheses around the argument. Example: \[\begin{align*} {\log}_7(4)\text{ is equivalent to } {\log}_74\end{align*}\nonumber\]
- Common logarithms are written as \(y=\log(x)\). All other logarithms, excluding the natural logarithm must have the base written or else it is assumed to be a common logarithm
- The natural logarithm is written as \(y=\ln(x)\). This is the only logarithm with a function name, \(ln\), that is different than all other logarithms.
Let's end with some examples of using the relationship between exponential and logarithmic functions to find inverse functions.
Given \(J(x)=\log_3(x-2)+4\), find \(J^{-1}\)
Solution
\[\begin{align*} J(x)&=\log_3(x-2)+4\\[4pt]y&=\log_3(x-2)+4\\[4pt]x&=\log_3(y-2)+4\\[4pt]x-4&=\log_3(y-2)\\[4pt]3^{x-4}&=y-2\\[4pt]3^{x-4}+2&=y\\ \end{align*}\]
Hence \(J^{-1}=3^{x-4}+2\)
Notice that \(J(x)\) is a basic logarithm function of base 3 that has a horizontal shift to the right 2 units and a vertical shift up 4 units. \(J^{-1}\) has a horizontal shift to the right 4 units and a vertical shift up 2 units. When working with transformed functions, the inverse will have its transformations reversed between the horizontal and vertical directions. Remember that the domain and range of functions and their inverse are also reversed.
Find the inverse of \(V(x)=5-e^{x+3}\)
- Answer
-
\(y=\ln(5-x)-3\)