3.4 What a logarithm is
Activity 3.4.2
For each of the following equations, determine the exact value of the unknown variable. If the exact value involves a logarithm, use a computational device to report an approximate value. For instance, if the exact value is [latex]y = log_{10}(2)[/latex], you can also note that [latex]y \approx 0.301.[/latex]
a. [latex]10^t = 0.00001[/latex]
b. [latex]\log_{10}(100000)[/latex]
c. [latex]10^t=37[/latex]
d. [latex]\log_{10}(y) = 1.375[/latex]
e. [latex]10^t = 0.04[/latex]
f. [latex]3 \cdot 10^t + 11 = 147[/latex]
g. [latex]2log_{10}(y)+5=1[/latex]
Show Solution
a. This expressions says the power we can raise [latex]10[/latex] to get [latex]0.00001[/latex]which is [latex]t=-5.[/latex]
b. This expressions says the power we can raise [latex]10[/latex] to get [latex]1000000[/latex] which is [latex]t=6.[/latex]
c. Rewrite the exponential function as its equivalent logarithm expression [latex]t = \log_{10}(37) \approx 1.5682.[/latex]
d. Rewrite the logarithmic expression as its equivalent exponential expression [latex]y = 10^{1.375} \approx 23.7137.[/latex]
e. Rewrite the exponential expression as its equivalent logarithmic expression
[latex]t = log_{10}(0.04) = -1.3979.[/latex]
f. Start by isolating the exponential term:
\begin{align*}
2\cdot 10^t + 11 &= 147 \\
2\cdot 10^t &= 136 \\
10^t &= 68 \\
\end{align*}
Rewrite [latex]10^t = 68[/latex] as its equivalent logarithmic expression to get [latex]t = log_{10}(68) \approx 1.8325.[/latex]
g. Start by isolating the logarithmic term
\begin{align*}
2log_{10}(y) + 5 &= 1 \\
2log_{10}(y) &= -4 \\
log_{10}(y) &= -2 \\
\end{align*}
Rewrite [latex]log_{10}(y) = -2 [/latex] as its equivalent exponential expression
[latex]y = 10^{-2} = \frac{1}{10^2} = \frac{1}{100}.[/latex]
Activity 3.4.4
Solve each of the following equations for the exact value of the unknown variable. If there is no solution to the equation, explain why not.
a. [latex]e^t = \frac{1}{10}[/latex]
b. [latex]5e^t = 7[/latex]
c. [latex]\ln(t) = -\frac{1}{3}[/latex]
d. [latex]e^{1-3t} = 4[/latex]
e. [latex]2\ln(t)+1=4[/latex]
f. [latex]4-3e^{2t}=2[/latex]
g. [latex]4+3e^{2t} = 2[/latex]
h. [latex]\ln(5-6t) = -2[/latex]
Show Solution
a. Rewrite [latex]e^{t} = \frac{1}{10}[/latex] as its equivalent logarithmic expression [latex]t = \ln(\frac{1}{10}).[/latex]
b. First isolate the exponential term by dividing both sides by 5 to get [latex]e^t = \frac{7}{5}[/latex]. Rewrite [latex]e^t = \frac{7}{5}[/latex] as its equivalent logarithmic expression to get [latex]t = \ln(\frac{7}{5}).[/latex]
c. Rewrite [latex]\ln(t) = \frac{-1}{3}[/latex] as its equivalent exponential expression [latex]t = e^{\frac{-1}{3}}.[/latex]
d. Rewrite [latex]e^{1-3t} = 4[/latex] as its equivalent logarithmic expression [latex]1-3t = \ln(4).[/latex] Then we will isolate [latex]t.[/latex]
\begin{align*}
1-3t &= \ln(4) \\
-3t &= \ln(4) -1 \\
t &= \frac{\ln(4)-1}{-3} \\
\end{align*}
e. Isolate the natural log term
\begin{align*}
2\ln(t) + 1 &= 4 \\
2\ln(t) &= 3 \\
\ln(t) &= \frac{3}{2} \\
\end{align*}
Rewrite [latex]\ln(t) = \frac{3}{2}[/latex] as its equivalent exponential expression [latex]t = e^{\frac{3}{2}}.[/latex]
f. Isolate the exponential term
\begin{align*}
4-3e^{2t} &= 2 \\
-3e^{2t} &= -2 \\
e^{2t} &= \frac{2}{3} \\
\end{align*}
Rewrite [latex]e^{2t} = \frac{2}{3}[/latex] as its equivalent logarithmic expression [latex]2t = \ln(\frac{2}{3})[/latex], then isolate t by dividing both sides by 2 to get [latex]t = \frac{\ln(\frac{2}{3})}{2}.[/latex]
g. Start by isolating the exponential term
\begin{align*}
4+3e^{2t} &= 2 \\
3e^{2t} &= -2 \\
e^{2t} &= \frac{-2}{3} \\
\end{align*}
Rewrite [latex]e^{2t} = \frac{-2}{3}[/latex] as its equivalent logarithmic expression [latex]2t = \ln(\frac{-2}{3})[/latex], then isolate [latex]t[/latex] by dividing both sides by $2$ to get [latex]t =\frac{\ln(\frac{-2}{3})}{2}[/latex]. However, the domain of logarithmic functions are [latex](0,\infty)[/latex]. So our answer is not in the domain so there is no solution.
h. Rewrite \[latex]ln(5-6t) = -2[/latex]as its equivalent exponential expression [latex]5-6t = e^{-2}[/latex]. Then we isolate [latex]t[/latex]
\begin{align*}
5-6t &= e^{-2} \\
-6t &= e^{-2} – 5 \\
t &= \frac{e^{-2} – 5}{-6} \\
\end{align*}