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.

E 3.4 Exercises