Exponential equations I

Some exponential equations can be solved by matching bases. When that is not possible (because the right-hand side is not a nice power), we use logarithms to bring the unknown index down.

Technique: equating indices

If both sides of an equation can be expressed with the same base, we can equate the indices:

For example, $2^{x} = 32 = 2^{5} ⟹ x = 5.$

Technique: the log power rule

For any base $a > 0,$ $a \neq 1$ :

$\log_{a} x^{n} = n \log_{a} x$

This lets us move an unknown index out of the exponent position. For example, $\lg 2^{x} = x \lg 2.$

Technique: solving with logarithms

When the right-hand side is not a convenient power of the base, take logarithms of both sides.

General base — for $b^{x} = k,$ take $\lg$ (or $\ln$ ) of both sides:

$\begin{matrix} b^{x} & = k \\ \lg b^{x} & = \lg k \\ x \lg b & = \lg k \\ x & = \frac{\lg k}{\lg b} \end{matrix}$

Base $e$ — for $e^{x} = k,$ take $\ln$ of both sides:

$\begin{matrix} e^{x} & = k \\ \ln e^{x} & = \ln k \\ x & = \ln k \end{matrix}$

Example: Exponential equations

Question

Solve

$3^{x} = 81$
$2^{x} = 81$
$e^{x} = 5$
$3 + 2 e^{4 x - 1} = 9$

Solution

Express the right-hand side as a power of 3:

$\begin{matrix} 3^{x} & = 3^{4} \\ x & = 4 ∎ \end{matrix}$

81 is not a power of 2, so take $\lg$ of both sides:

$\begin{matrix} x \lg 2 & = \lg 81 \\ x & = \frac{\lg 81}{\lg 2} \\ x & \approx 6.34 ∎ \end{matrix}$

Take $\ln$ of both sides:

$\begin{matrix} e^{x} & = 5 \\ \ln e^{x} & = \ln 5 \\ x & = \ln 5 \\ x & \approx 1.61 ∎ \end{matrix}$

Isolate the exponential term first:

$\begin{matrix} 3 + 2 e^{4 x - 1} & = 9 \\ 2 e^{4 x - 1} & = 6 \\ e^{4 x - 1} & = 3 \end{matrix}$

Take $\ln$ of both sides:

$\begin{matrix} \ln e^{4 x - 1} & = \ln 3 \\ 4 x - 1 & = \ln 3 \\ 4 x & = 1 + \ln 3 \\ x & = \frac{1 + \ln 3}{4} \\ x & \approx 0.525 ∎ \end{matrix}$