For \(a>0\) and \(a\ne1\):
\(a^u=a^v\quad\Longleftrightarrow\quad u=v\)
Methods
| Form | Method | Example |
|---|---|---|
| Powers of the same base | Equate indices. | \(2^{3x-1}=8\Rightarrow3x-1=3\Rightarrow x=\frac43\) |
| Quadratic in \(a^x\) | Let \(u=a^x>0\). | Solve for u, reject non-positive values, then find x. |
Useful substitutions
| Equation | Substitution | Because |
|---|---|---|
| \(9^x-4(3^x)+3=0\) | ||
| \(\mathrm{e}^{2x}-7\mathrm{e}^{x}+10=0\) | ||
| \(\mathrm{e}^{4x}-3\mathrm{e}^{2x}-4=0\) | ||
| \(25^x=4-3(25^{-x})\) |
Factor out shifted powers first, e.g. \(2^{x+1}=2(2^x)\).
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