Solving exponential equations by substitution

An equation that contains \(a^{2x}\), \(a^{x+1}\) or \(a^{-x}\) together with \(a^x\) is a quadratic equation in disguise.

Method

1. Write every power in terms of \(a^x\): \(a^{2x}=(a^x)^2\), \(a^{x+1}=a\,(a^x)\), \(a^{-x}=\dfrac{1}{a^x}\).
2. Let \(u=a^x\), where \(u>0\), and solve the quadratic equation in \(u\).
3. Reject any value of \(u\) that is zero or negative, because \(a^x\) is always positive.
4. Solve \(a^x=u\) for each remaining value by writing \(u\) as a power of \(a\).

Example

Solve \(4^x+2^{x+1}=8\).

\(\begin{aligned} (2^x)^2+2(2^x)-8&=0 \\ u^2+2u-8&=0\qquad(u=2^x) \\ (u+4)(u-2)&=0 \end{aligned}\)

\(u=2\) or \(u=-4\) (rejected, since \(2^x>0\)).

\(2^x=2=2^1\), so \(x=1\).