3.4 | Exponential and Logarithmic Equations

To solve an exponential equation of the form
$$a b^{k x -c} + d = 0$$

• Isolate the exponential.
• Take logarithm base $$b$$ of both sides. If you obtain an expression involving taking the logarithm of a negative number, then no solution exists. Otherwise continue to the next step.
• Use the inverse property to obtain a linear equation.
• Now solve for $$x$$.
Strategy for Solving Exponential Equations Given an exponential equation, rewrite the equation to obtain equations of the type
$$a b^{kx-c} +d = 0.$$
This may require factoring, combining fractions, etc…
Find the exact solution to
$$68e^{4x+1}+14 = 16.$$

• Isolate the exponential.
$$e^{4x+1} = \frac{1}{34}$$
• Take natural logarithm of both sides and apply the inverse property for logarithms.
\begin{aligned} \ln e^{4x+1} &= \ln(1/34) \\ 4x+1 & = \ln(1/34) \end{aligned}
• Solve for $$x$$.
$$x = \frac{\ln(1/34) -1}{4}$$
Solve
$$\left( 4 – \frac{2.471}{40} \right)^{9t} = 21.$$

\begin{aligned} \log \left( 4 – \frac{2.471}{40} \right)^{9t} & = \log 21 \\ 9t \log \left( 4 – \frac{2.471}{40} \right) & = \log 21 \\ t & = \frac{\log 21}{9 \log \left( 4 – \frac{2.471}{40} \right)} \\ t & \approx 0.246788 \end{aligned}
Solve
$$e^{2x} -2 e^{x} – 3 = 0$$

• Factor.
$$(e^x +1) (e^x-3) = 0$$
• Set each factor equal to zero.
$$e^x +1 = 0 \Rightarrow x = \ln(-1)$$
or
$$e^x – 3 = 0 \Rightarrow x = \ln 3$$

Thus, the only solution to the given equation is $$x = \ln 3$$ since $$\ln(-1)$$ is undefined.

To solve an logarithmic equation of the form
$$a \log_b (k x -c) + d = 0$$

• Consolidate and Isolate the logarithms.
• Exponentiate both sides base $$b$$.
• Use the inverse property to obtain a linear equation.
• Now solve for $$x$$.
Strategy for Solving Logarithmic Equations Given a logarithmic equation, rewrite the equation to obtain equations of the type
$$a \log_b (k x -c) + d = 0.$$
This may require factoring, combining fractions, etc…
Find the exact solution to
$$-2 +2 \ln(3x) = 17.$$

• Isolate the logarithm.
$$\ln(3x) = \frac{19}{2}$$
• Exponentiate both sides base $$e$$ and apply the inverse property.
\begin{aligned} e^{\ln(3x)} &= e^{19/2} \\ 3x &= e^{19/2} \end{aligned}
• Solve for $$x$$.
$$x = \frac{1}{3} e^{19/2}$$
Solve
$$\log_3(x+5) = \log_3(x-1) – \log_3(x+1)$$

• Consolidate and Isolate the logarithm.
$$\log_3\left( \frac{x+5}{x-1} \cdot (x+1) \right) = 0$$
• Exponentiate both sides base $$3$$ and apply the inverse property.
\begin{aligned} 3^{\log_3\left( \frac{x+5}{x-1} \cdot (x+1) \right)} &= 3^0 \\ \frac{x+5}{x-1} \cdot (x+1) &= 1 \end{aligned}
• Solve for $$x$$.
\begin{aligned} (x+5)(x+1) & = x-1 \\ x^2 +6x+5 & = x-1 \\ x^2 +5x +6 & = 0 \\ (x +2)(x+3) & = 0 \end{aligned}
So,
$$x = -2 \text{ or } x =-3$$
• But $$x = -2,$$ and $$x = -3$$ both make the original equation undefined. Therefore, the given equation has no solutions.