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Exploring Differential Equations An Interactive, Student-Centered Approach

Geoffrey W. Cox, Ph.D.

Section Exponential and Logarithmic Functions

Exponential and logarithmic expressions appear constantly in differential equations. They arise in the solutions of linear equations, in the Laplace transform and its inverse, and in modeling exponential growth, decay, and oscillation. This review gathers the essential algebraic rules you’ll use throughout the course.
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✳️ Exponential Rules.

General Natural (\(a=e\))
\(E_1:\)
\begin{gather*} \ds (ab)^x = a^x \cdot b^x \end{gather*}
\begin{gather*} \ds (eb)^x = e^x \cdot b^x \end{gather*}
\(E_2:\)
\begin{gather*} \ds a^x \cdot a^y = a^{x+y} \end{gather*}
\begin{gather*} \ds e^x \cdot e^y = e^{x+y} \end{gather*}
\(E_3:\)
\begin{gather*} \ds (a^x)^y = a^{xy} \end{gather*}
\begin{gather*} \ds (e^x)^y = e^{xy} \end{gather*}
\(E_4:\)
\begin{gather*} \ds a^{-x} = \frac{1}{a^x} \end{gather*}
\begin{gather*} \ds e^{-x} = \frac{1}{e^x} \end{gather*}
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✳️ Logarithmic Rules.

General Natural (\(a=e\))
\(L_1:\)
\begin{gather*} \ds \log_b (b^x) = x \end{gather*}
\begin{gather*} \ds \ln(e^x) = x \end{gather*}
\(L_2:\)
\begin{gather*} \ds b^{\log_b(x)} = x \end{gather*}
\begin{gather*} \ds e^{\ln(x)} = x \end{gather*}
\(L_3:\)
\begin{gather*} \ds \log_b(1) = 0 \end{gather*}
\begin{gather*} \ds \ln(1) = 0 \end{gather*}
\(L_4:\)
\begin{gather*} \ds \log_b(xy) = \log_b(x) + \log_b(y) \end{gather*}
\begin{gather*} \ds \ln(xy) = \ln(x) + \ln(y) \end{gather*}
\(L_5:\)
\begin{gather*} \ds \log_b\left(\frac{\ds x}{\ds y}\right) = \log_b(x) - \log_b(y) \end{gather*}
\begin{gather*} \ds \ln\left(\frac{\ds x}{\ds y}\right) = \ln(x) - \ln(y) \end{gather*}
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These identities will help you solve equations involving exponentials or logarithmsβ€”common in modeling growth and decay, solving linear ODEs, and interpreting Laplace transforms.
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🌌 Example 309.

Solve for \(x\text{:}\) \(\ds \quad e^{3x+z} - e^z = y\)
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Solution.
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*}
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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.
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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.
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🌌 Example 310.

Solve for \(x\text{:}\) \(\ds \quad \ln(x+y) = 5 + \ln(z)\)
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Solution.
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*}
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Now you try. Solve for \(x\) in each of the following equations using the rules above.
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  1. \(e^{x + y} = 12\)
    Solution.
    \begin{gather*} x = \ln(12) - y \end{gather*}
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    Answer.
    \begin{equation*} x = \ln(12) - y \end{equation*}
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  2. \(e^{x + y} + e^x = 12\)
    Solution.
    \begin{gather*} e^x(e^y + 1) = 12\\ e^x = \frac{12}{e^y + 1}\\ x = \ln(12) - \ln(e^y + 1) \end{gather*}
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    Answer.
    \begin{equation*} x = \ln\left( \frac{12}{e^y + 1} \right) \end{equation*}
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  3. \(e^x = 1\)
    Solution.
    \begin{gather*} x = \ln(1) = 0 \end{gather*}
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    Answer.
    \begin{equation*} x = 0 \end{equation*}
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  4. \(\ln x = 3 \ln z\)
    Solution.
    \begin{gather*} x = e^{\ln(z^3)} = z^3 \end{gather*}
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    Answer.
    \begin{equation*} x = z^3 \end{equation*}
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  5. \(y + \ln x = 4\)
    Solution.
    \begin{gather*} x = e^{4 - y} \end{gather*}
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    Answer.
    \begin{equation*} x = e^{4 - y} \end{equation*}
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  6. \(\ln y + \ln x = 4\)
    Solution.
    \begin{gather*} \ln(xy) = 4\\ xy = e^4\\ x = \frac{e^4}{y} \end{gather*}
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    Answer.
    \begin{equation*} x = \frac{e^4}{y} \end{equation*}
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  7. \(\ln x = 8 \ln y + 5\)
    Solution.
    \begin{gather*} x = e^{5} \cdot e^{\ln(y^8)} = e^5 y^8 \end{gather*}
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    Answer.
    \begin{equation*} x = e^5 y^8 \end{equation*}
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πŸ“™ Definition 311. Euler’s Formula.

Euler’s formula relates complex exponentials to trigonometric functions. For any real number \(\theta\text{:}\)
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\begin{equation*} e^{i\theta} = \cos(\theta) + i\sin(\theta) \end{equation*}
As a result, we also have:
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\begin{equation*} e^{a + ib} = e^a \cdot e^{ib} = e^a(\cos(b) + i\sin(b)) \end{equation*}
This identity is fundamental in solving linear differential equations with complex roots.
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