The two most useful techniques for solving equations involving exponentials are:

The first enables you to manipulate terms in the equation using the rules $a^x{a}^y\=a^{x\+y}$,  $\frac{a^x}{a^y}\=a^{x-y}$ and ${\left(a^x\right)}^y\=a^{x\cdot y}$. The second lets you take advantage of the rule $a^x{b}^x\={\left(a\cdot b\right)}^x$. It is often the case that the two rules are used in tandem; for example, if $c\=a\cdot b$ then $\frac{a^x}{c^x}\=\frac{a^x}{{\left(a\cdot b\right)}^x}\=\frac{a^x}{a^x{b}^x}\=b^{-x}$.

Example: Find $x$ such that $\frac{8^{2 x}}{4^{5 x}}\={16}^{}$.

Solution: Note that 8 and 4 can both be expressed as powers of 2 ($8\=2^3$ and $4\=2^2\)$, so $\frac{8^{2 x}}{4^{5 x}}\=\frac{{\left(2^3\right)}^{2 x}}{{\left(2^2\right)}^{5 x}}\=\frac{2^{6 x}}{2^{10 x}}\=2^{-4 x}\={\left(2^4\right)}^{-x}\={16}^{-x}\=16$, so $x\=-1$.

See the lessons on logarithms elsewhere in this collection for information regarding the properties of logarithms.

An equation which can be recast into the form $a^x\=b$ can be solved for $x$ in terms of logarithms: $x\={\log }_a\left(b\right)$.

Example: Find $x$ such that $3^{3 x}\=\frac{17}{9^x}$.

Solution: Since $9\=3^2$ we have $3^{3 x}\=\frac{17}{9^x}\=\frac{17}{{\left(3^2\right)}^x}\=\frac{17}{3^{2 x}}$. Multiplying through by $3^{2 x}$, we get

$3^{3 x}\cdot 3^{2 x}\=3^{\left(3 x\+2 x\right)}\=3^{5 x}\=17$ 

Taking the base-3 logarithm of both sides, we have $5 x\={\log }_3\left(17\right)$, so, finally, $x\=\frac{1}{5}{\log }_3\left(17\right)$.