Section A.1 Exponential and Logarithmic Functions
Recall the following rules for exponential and logarithmic functions.
Exponential Rules.
|
General |
|
Natural (\(a=e\)) |
| \(E_1:\) |
\begin{gather*}
\displaystyle (ab)^x = a^x \cdot b^x
\end{gather*}
|
|
\begin{gather*}
\displaystyle (eb)^x = e^x \cdot b^x
\end{gather*}
|
| \(E_2:\) |
\begin{gather*}
\displaystyle a^x \cdot a^y = a^{x+y}
\end{gather*}
|
|
\begin{gather*}
\displaystyle e^x \cdot e^y = e^{x+y}
\end{gather*}
|
| \(E_3:\) |
\begin{gather*}
\displaystyle (a^x)^y = a^{xy}
\end{gather*}
|
|
\begin{gather*}
\displaystyle (e^x)^y = e^{xy}
\end{gather*}
|
| \(E_4:\) |
\begin{gather*}
\displaystyle a^{-x} = \frac{1}{a^x}
\end{gather*}
|
|
\begin{gather*}
\displaystyle e^{-x} = \frac{1}{e^x}
\end{gather*}
|
Logarithmic Rules.
|
General |
|
Natural (\(a=e\)) |
| \(L_1:\) |
\begin{gather*}
\displaystyle \log_b (b^x) = x
\end{gather*}
|
|
\begin{gather*}
\displaystyle \ln(e^x) = x
\end{gather*}
|
| \(L_2:\) |
\begin{gather*}
\displaystyle b^{\log_b(x)} = x
\end{gather*}
|
|
\begin{gather*}
\displaystyle e^{\ln(x)} = x
\end{gather*}
|
| \(L_3:\) |
\begin{gather*}
\displaystyle \log_b(1) = 0
\end{gather*}
|
|
\begin{gather*}
\displaystyle \ln(1) = 0
\end{gather*}
|
| \(L_4:\) |
\begin{gather*}
\displaystyle \log_b(xy) = \log_b(x) + \log_b(y)
\end{gather*}
|
|
\begin{gather*}
\displaystyle \ln(xy) = \ln(x) + \ln(y)
\end{gather*}
|
| \(L_5:\) |
\begin{gather*}
\displaystyle \log_b\left(\frac{\displaystyle x}{\displaystyle y}\right) = \log_b(x) - \log_b(y)
\end{gather*}
|
|
\begin{gather*}
\displaystyle \ln\left(\frac{\displaystyle x}{\displaystyle y}\right) = \ln(x) - \ln(y)
\end{gather*}
|
Let’s look at a couple of examples, starting with an equation containing exponentials.
Example 53.
Solve for \(x\text{:}\) \(\displaystyle \quad e^{3x+z} - e^z = y\)
Solution.
If you prefer to see a video of this example,
click here
We might begin by isolating the exponential that contains \(x\) and then taking the natural log of both sides.
\begin{align*}
e^{3x+z} - e^z =\amp\ y\\
e^{3x+z} =\amp\ y + e^z\\
\knowl{./knowl/xref/ln_rule_01.html}{\text{\(\overset{L_1}{\hookrightarrow}\qquad\)}}
\ln\big(e^{3x+z}\big) =\amp\ \ln\big(y + e^z\big)\\
3x + z =\amp\ \ln\big(y + e^z\big)\\
3x =\amp\ \ln\big(y + e^z\big) - z\\
x =\amp\ \frac{1}{3}\ln\big(y + e^z\big) - \frac{1}{3}z
\end{align*}
It’s worth noting that we cannot break up that log on the right hand side. There’s no "rule" that helps when we have addition inside a logarithm.
There is another way to approach this if notice that \(z\) appears inside both exponential terms.
\begin{align*}
\knowl{./knowl/xref/exp_rule_02e.html}{\text{\(\overset{E_2}{\hookrightarrow}\qquad\)}}
e^{3x+z} - e^z =\amp\ y\\
e^{3x}\cdot e^{z} - e^z =\amp\ y\\
e^z(e^{3x} - 1) =\amp\ y\\
e^{3x} - 1 =\amp\ \frac{y}{e^z}
\knowl{./knowl/xref/exp_rule_04e.html}{\text{\(\qquad\overset{E_4}{\hookleftarrow}\)}}\\
e^{3x} - 1 =\amp\ ye^{-z}\\
e^{3x} =\amp\ ye^{-z} + 1\\
\knowl{./knowl/xref/ln_rule_01.html}{\text{\(\overset{L_1}{\hookrightarrow}\qquad\)}}
\ln\big( e^{3x} \big) =\amp\ \ln \big( ye^{-z} + 1\big)\\
3x =\amp\ \ln \big( ye^{-z} + 1 \big) \\
x =\amp\ \frac{1}{3}\ln\big(ye^{-z}+1\big)
\end{align*}
The answers may look different, but they are equivalent and both are correct.
Now let’s look at an example involving logarithms.
Example 54.
Solve for \(x\text{:}\) \(\displaystyle \quad \ln(x+y) = 5 + \ln(z)\)
Solution.
If you prefer to see a video of this example,
click here
We’ll carefully apply the rules above. We want to get our hands on \(x\text{,}\) and right now its inside a logarithm. In order to undo that, we’ll exponentiate both sides.
\begin{align*}
\ln(x+y) =\amp\ 5 + \ln(z)\\
\knowl{./knowl/xref/ln_rule_02.html}{\text{\(\overset{L_2}{\hookrightarrow}\qquad\)}}
e^{\ln(x+y)} =\amp\ e^{\big(5 + \ln(z)\big)}
\knowl{./knowl/xref/exp_rule_02e.html}{\text{\(\qquad\overset{E_2}{\hookleftarrow}\)}}\\
x + y =\amp\ e^{5} \cdot e^{\ln(z)}
\knowl{./knowl/xref/ln_rule_02.html}{\text{\(\qquad\overset{L_2}{\hookleftarrow}\)}}\\
x =\amp\ e^{5}z - y
\end{align*}
Now you should try. Be careful!
Use algebra and the rules above to solve for \(x\) in each of the following equations.
\(e^{x+y} = 12\) Solution.
\begin{align*}
e^{x+y} =\amp\ 12 \\
\ln(e^{x+y}) =\amp\ \ln(12) \\
x+y =\amp\ \ln(12) \\
x =\amp\ \ln(12) - y
\end{align*}
Answer.
\begin{equation*}
x = \ln(12) - y
\end{equation*}
\(e^{x+y} + e^x= 12\) Solution.
\begin{align*}
e^{x+y} + e^x =\amp\ 12 \\
e^x\cdot e^y + e^x =\amp\ 12 \\
e^x(e^y + 1) =\amp\ 12 \\
e^x =\amp\ \frac{12}{e^y + 1} \\
\ln(e^x) =\amp\ \ln \left( \frac{12}{e^y + 1} \right) \\
x =\amp\ \ln \left( \frac{12}{e^y + 1} \right),\qquad \text{or} \\
=\amp\ \ln(12) - \ln(e^y + 1)
\end{align*}
Answer.
\begin{align*}
x =\amp\ \ln\left( \frac{12}{e^y + 1} \right), \qquad \text{or}\\
=\amp\ \ln(12) - \ln(e^y + 1)
\end{align*}
\(e^x = 1 \qquad\) Solution.
\begin{align*}
e^x =\amp\ 1 \\
\ln(e^x) =\amp\ \ln(1) \\
x =\amp\ 0
\end{align*}
Answer.
\begin{equation*}
x = 0
\end{equation*}
\(\ln x = 3\ln z \qquad\) Solution.
\begin{align*}
\ln x =\amp\ 3\ln z \\
e^{\ln x} =\amp\ e^{3\ln z} \\
e^{\ln x} =\amp\ e^{\ln (z^3)} \\
x =\amp\ z^3
\end{align*}
Answer.
\begin{equation*}
x = z^3
\end{equation*}
\(y + \ln x = 4 \qquad\) Solution.
\begin{align*}
y + \ln x =\amp\ 4 \\
\ln x =\amp\ 4 - y \\
e^{\ln x}=\amp\ e^{4 - y} \\
x =\amp\ e^{4-y}
\end{align*}
Answer.
\begin{equation*}
x = e^{4-y}
\end{equation*}
\(\ln y + \ln x = 4 \qquad\) Solution.
\begin{align*}
\ln y + \ln x =\amp\ 4 \\
\ln x =\amp\ 4 - \ln y \\
e^{\ln x} =\amp\ e^{4 - \ln y} \\
x =\amp\ e^{4 - \ln y} \\
=\amp\ e^4 \cdot e^{-\ln y} \\
=\amp\ e^4 \cdot e^{\ln (y^{-1})} \\
=\amp\ e^4 \cdot y^{-1} \\
=\amp\ e^4 \cdot \frac{1}{y} \\
=\amp\ \frac{e^4}{y}
\end{align*}
Answer.
\begin{equation*}
x = \frac{e^4}{y}
\end{equation*}
\(\ln x = 8\ln y + 5 \qquad\) Solution.
\begin{align*}
\ln x =\amp\ 8\ln y + 5 \\
e^{\ln x} =\amp\ e^{8\ln y + 5} \\
x =\amp\ e^{8\ln y}\cdot e^5 \\
=\amp\ e^{\ln (y^8)}\cdot e^5 \\
=\amp\ y^8\cdot e^5 \\
=\amp\ e^5 y^8
\end{align*}
Answer.
\begin{equation*}
x = e^5 y^8
\end{equation*}
Definition 55. Euler’s Formula.
You have attempted
of
activities on this page.
