Exponential equations II: Substitution

Some exponential equations can be transformed into quadratic equations using a substitution.

Technique: substitution

If an equation contains terms like $a^{2 x}$ and $a^{x},$ we can use the substitution $u = a^{x} .$

Since $a^{2 x} = (a^{x})^{2} = u^{2},$ the equation becomes a quadratic in $u .$

Note: Since $a^{x} > 0$ for all real $x,$ we must have $u > 0.$ Any negative or zero values for $u$ should be rejected.

For equations with $e^{x}$ and $e^{- x},$ use $u = e^{x}$ so that $e^{- x} = \frac{1}{e^{x}} = \frac{1}{u},$ then multiply through by $u$ to clear the fraction.

Example A

Question A

Solve the equation $2^{2 x} - 3 (2^{x + 1}) + 8 = 0.$

Solution A

$\begin{matrix} 2^{2 x} - 3 (2^{x + 1}) + 8 & = 0 \\ 2^{2 x} - 3 (2^{x} \cdot 2^{1}) + 8 & = 0 \\ 2^{2 x} - 6 (2^{x}) + 8 & = 0 \end{matrix}$

Let $u = 2^{x} .$ The equation becomes:

$\begin{matrix} u^{2} - 6 u + 8 & = 0 \\ (u - 2) (u - 4) & = 0 \end{matrix}$

Since both values are positive, we solve for $x$ :

$\begin{matrix} u & = 2 & or & u & = 4 \\ 2^{x} & = 2 & or & 2^{x} & = 4 \\ 2^{x} & = 2^{1} & or & 2^{x} & = 2^{2} \\ x & = 1 ∎ & or & x & = 2 ∎ \end{matrix}$

Example B

Question B

Solve the equation $2 e^{x} - 5 + 2 e^{- x} = 0.$

Solution B

Let $u = e^{x},$ so $e^{- x} = \frac{1}{u} .$ Multiply through by $u$ :

$\begin{matrix} 2 u - 5 + \frac{2}{u} & = 0 \\ 2 u^{2} - 5 u + 2 & = 0 \\ (2 u - 1) (u - 2) & = 0 \end{matrix}$

Since both values are positive, we solve for $x$ :

$\begin{matrix} e^{x} & = \frac{1}{2} & or & e^{x} & = 2 \\ x & = \ln \frac{1}{2} & x & = \ln 2 \\ x & \approx - 0.693 & x & \approx 0.693 ∎ \end{matrix}$