Example 6.4.1.
Solve
\begin{equation*}
\log_3(x)=2\text{.}
\end{equation*}
We can re-write this as
\begin{equation*}
3^2=x
\end{equation*}
which tells us that \(x=9\text{.}\)
Section 6.4 Solving Equations Using Logarithm
We can evaluate expressions with exponentials and logarithms using devices. We can also solve equations involving logarithmic and exponential expressions. We will use both the definition of logarithm and the relationship between logarithmic and exponential expressions.
Subsection 6.4.1 Evaluating and Solving with Logs
First, we will use Definition 6.3.2 to solve logarithmic equations.
Example 6.4.2.
Solve
\begin{equation*}
\log_2(x+5)=7\text{.}
\end{equation*}
We can re-write this as
\begin{align*}
2^7 \amp = x+5.\\
128 \amp = x+5.\\
123 \amp = x.
\end{align*}
Checkpoint 6.4.3.
Solve \({5}\log_{{5}}(x)={15}\text{.}\)
Answer.
\(125\)
Solution.
\begin{equation*}
\begin{aligned}
{5}\log_{{5}}(x) \amp = {15}\\
{5}\log_{{5}}(x)/{5} \amp = {15}/{5}\\
\log_{{5}}(x) \amp = {3}\\
{5}^{{3}} \amp = x\\
{125} \amp = x
\end{aligned}
\end{equation*}
Subsection 6.4.2 Additional Logarithm Property
We need one more property of logarithms to solve some equations.
Example 6.4.4.
Evaluate \(\log(10^7)\)
Solution.
\(\log(10^7)=7 \log(10) = 7\) because \(\log(10)\) is the same as \(10^x=10\) so \(x=1\text{.}\)
In general the property is
\begin{equation*}
\log_b(a^p) = p \log_b(a)\text{.}
\end{equation*}
We can use this property to solve equations involving exponentials.
Example 6.4.5.
Solve \(5 = 2^x\text{.}\)
Solution.
Because the exponential is base 2, we will use a log base 2.
\begin{align*}
5 & = 2^x.\\
\log_2(5) & = \log_2(2^x).\\
\log_2(5) & = x\log_2(2).\\
\log_2(5) & = x \cdot 1.\\
\log_2(5) & = x.
\end{align*}
Note those last three steps will always be the same, so we often jump from the second step to the last.
Your calculator likely does not have a button for calculating \(\log_2(x)\text{.}\) The property we used does not depend on using that log however.
Example 6.4.6.
Solve \(5 = 2^x\text{.}\)
Solution.
Because the exponential is base 2, we will use a log base 2.
\begin{align*}
5 & = 2^x.\\
\ln(5) & = \ln(2^x).\\
\ln(5) & = x\ln(2).\\
\frac{\ln(5)}{\ln(2)} & = \frac{x\ln(2)}{\ln(2)}.\\
\frac{\ln(5)}{\ln(2)} & = x.
\end{align*}
Your calculator does have a button for \(\ln(x)\text{.}\) Note this implies that \(\log_2(5) = \frac{\ln(5)}{\ln(2)}\text{.}\) This relationship always works.
Example 6.4.7.
Solve \(13 = 2+3e^{x-1}\text{.}\)
Solution.
Order of operations dictates the following
\begin{align*}
13 & = 2+3e^{x-1}.\\
11 & = 3e^{x-1}.\\
\frac{11}{3} & = e^{x-1}\\
\ln\left(\frac{11}{3}\right) & = \ln\left(e^{x-1}\right).\\
\ln\left(\frac{11}{3}\right) & = x-1.\\
\ln\left(\frac{11}{3}\right)+1 & = x.\\
2.299 & \approx x.
\end{align*}
We can use the relationship between exponentials and logarithms to solve logarithmic equations as well.
Example 6.4.8.
Solve \(14 = 12+\log(5x)\text{.}\)
Solution.
First we combine the terms outside the logarithm, the we re-write that as an exponential.
\begin{align*}
14 & = 12+\log(5x)\\
2 & = \log(5x)\\
10^2 & = 5x\\
100 & = 5x\\
20 & \approx x
\end{align*}
Example 6.4.9.
Solve \(7 = \ln(3x+2)\text{.}\)
Solution.
We re-write this as an exponential.
\begin{align*}
7 & = \ln(3x+2)\\
e^7 & = 3x+2\\
e^7-2 & = 3x\\
\frac{1}{3}(e^7-2) & = x\\
365 & \approx x
\end{align*}
Checkpoint 6.4.10.
Solve
\begin{equation*}
{5} = {5}\ln(x+{5})
\end{equation*}
\(x=\) .
Precision should be at least four decimal places.
Answer.
\(-2.28172\)
Subsection 6.4.3 Applications with Exponentials
Example 6.4.11.
Plutonium-241 has a half-life of 14.4 years. This means if you start with 10g of Pu-241 in 14.4 years there will be only 5g of Pu-241. Generally, this can be modeled by
\begin{equation*}
P=P_0 e^{kt}\text{.}
\end{equation*}
\(P_0\) is the initial amount of material. \(k\) is a constant that indicates how fast the material decays. \(P\) is the amount left after \(t\) units of time.
(a)
Write the model for Plutonium-241.
Solution.
We must first find \(k\text{.}\) We can use Example 6.4.7.
\begin{align*}
5 & = 10 e^{k(14.4)}\\
\frac{1}{2} & = e^{k(14.4)}\\
\ln\left(\frac{1}{2}\right) & = \ln\left(e^{k(14.4)}\right)\\
\ln\left(\frac{1}{2}\right) & = k(14.4).\\
\frac{1}{14.4}\ln\left(\frac{1}{2}\right) & = k\\
-0.0481 & \approx k.
\end{align*}
Thus the model is
\begin{equation*}
P = P_0 e^{-0.0481t}\text{.}
\end{equation*}
(b)
If a lab has 12g of Pu-241, how much will be left in 6 years?
Solution.
We use the model from the previous step.
\begin{align*}
P & = 12e^{-0.0481(6)}\\
& = 8.99.
\end{align*}
Example 6.4.12.
In Subsection 6.1.3 we produced the model \(P=4000 \cdot 2^{t/70}\text{.}\) Here we will redo this problem using the model \(P=P_0 e^{kt}\text{.}\)
The bacteria lactobacilus acidophilus doubles in population every 70 minutes. If the initial population was 4000 bacteria, what would the population be after 24 hours?
Solution.
First we calculate the constant \(k\) in the model.
\begin{align*}
8000 & = 4000e^{70 k}\\
\frac{8000}{4000} & = e^{70k}\\
2 & = e^{70k}\\
\ln(2) & = 70k\\
\frac{\ln(2)}{70} & = k\\
k & \approx 0.009902
\end{align*}
Using this value we can now calculate the value after 24 hours. Note 24 hours is \(24 \cdot 60 = 1440\) minutes.
\begin{align*}
P & = 4000e^{(1440)(0.009902)}\\
& \approx 6.232 \times 10^9
\end{align*}
Note, that the example in Subsection 6.1.3 and Example 6.4.12 imply that \(4000 \cdot 2^{t/70} = 4000 e^{(0.009902)t}\text{.}\) Generally, we can write \(2^x = e^{kx}\) or \(3^x = e^{kx}\) or similar for value of \(k\text{.}\)
Example 6.4.13.
Write \(2^x\) as \(e^{kx}\text{.}\)
Solution.
\begin{align*}
2^x & = e^{kx}.\\
\ln(2^x) & = \ln(e^{kx}).\\
x \ln(2) & = (kx)\ln(e).\\
x \ln(2) & = kx.\\
\frac{x \ln(2)}{x} & = \frac{kx}{x}.\\
\ln(2) & = k
\end{align*}
Thus \(2^x = e^{x \ln(2)}\text{.}\)
Example 6.4.14.
Acidity is measured in pH (percent hydrogen). It uses a logarithmic scale.
\begin{equation*}
\text{pH} = -\log(H_3O^+)
\end{equation*}
where \(H_3O^+\) is the concentration of hydronium ions per mole. This is obtained experimentally.
(a)
Hydrochloric acid has a concentration of 0.0025. Find its pH.
Solution.
\begin{align*}
\text{pH} & = -\log(0.0025)\\
& \approx 2.60.
\end{align*}
(b)
Sweat has a pH between 4.5 and 7. Suppose sweat is measured to have a pH of 5.3. Determine the concentration of ions.
Solution.
We setup the pH calculation and solve using Example 6.4.9.
\begin{align*}
5.3 & = -\log(c)\\
-5.3 & = \log(c)\\
10^{-5.3} & = 10^{\log(c)}\\
10^{-5.3} & = c\\
5.01 \times 10^{-6} & \approx c
\end{align*}
Example 6.4.15.
Larger earthquakes today are measured and reported using the moment magnitude scale. This is calculated via
\begin{equation*}
M_w = \frac{2}{3}\log(M_0)-10.7
\end{equation*}
where \(M_0\) is the seismic moment in Newtons per meter (a measure of energy).
(a)
Based on seismic readings \(M_0 = 7.2 \times 10^{22}\text{.}\) What was the moment magnitude?
Solution.
Using the formula we obtain
\begin{align*}
M_w & = \frac{2}{3}\log(7.2 \times 10^{22})\\
& \approx 4.5
\end{align*}
(b)
What was the seismic moment for an earthquake with magnitude 7.1?
Solution.
We setup the calculation and solve using Example 6.4.9.
\begin{align*}
7.1 & = \frac{2}{3}\log(M_0)-10.7\\
17.8 & = \frac{2}{3}\log(M_0)\\
26.7 & = \log(M_0)\\
10^{26.7} & = 10^{\log(M_0)}\\
5.01 \times 10^{26} & \approx M_0
\end{align*}
Exercises 6.4.4 Exercises
Exercise Group.
Solve equations with logarithms and exponentials
1. Contextless.
2. Contextless.
3. Contextless.
4. Contextless.
5. Contextless.
6. Contextless.
7. Contextless.
8. Contextless.
9. Contextless.
Exercise Group.
Use logarithms and exponentials to work with applications.
