# Interactive Differential Equations: A Step-by-Step Approach to Methods & Modeling

## SectionA.1Exponential 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.

## Example45.

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
1
www.youtube.com/watch?v=QdZekQ5_T2E
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.

## Example46.

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
2
www.youtube.com/watch?v=7xG6SisdjYE
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.
1. $$e^{x+y} = 12$$
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*}
\begin{equation*} x = \ln(12) - y \end{equation*}
2. $$e^{x+y} + e^x= 12$$
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*}
\begin{align*} x =\amp\ \ln\left( \frac{12}{e^y + 1} \right), \qquad \text{or}\\ =\amp\ \ln(12) - \ln(e^y + 1) \end{align*}
3. $$e^x = 1 \qquad$$
Solution.
\begin{align*} e^x =\amp\ 1 \\ \ln(e^x) =\amp\ \ln(1) \\ x =\amp\ 0 \end{align*}
\begin{equation*} x = 0 \end{equation*}
4. $$\ln x = 3\ln z \qquad$$
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*}
\begin{equation*} x = z^3 \end{equation*}
5. $$y + \ln x = 4 \qquad$$
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*}
\begin{equation*} x = e^{4-y} \end{equation*}
6. $$\ln y + \ln x = 4 \qquad$$
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*}
7. $$\ln x = 8\ln y + 5 \qquad$$