## College Algebra (10th Edition)

The solution set is $\left\{6\right\}$.
To solve the given equation, make the two sides have the same base. Note that $9 = 3^2$ and $27=3^3$, so the given equation is equivalent to: $(3^2)^{-x+15}=(3^3)^x$ Use the rule $(a^m)^n = a^{mn}$ to obtain: $3^{2(-x+15)} = 3^{3x} \\3^{-2x+30}=3^{3x}$ Use the rule $a^m=a^n \longrightarrow m=n$ to obtain: $-2x+30=3x$ Add $2x$ to both sides of the equation to obtain: $\begin{array}{ccc} &-2x+30+2x &= &3x+2x \\&30 &= &5x \end{array}$ Divide by 5 on both sides of the equation to obtain: $\begin{array}{ccc} &\frac{30}{5} &= &\frac{5x}{5} \\&6 &= &x \end{array}$ Thus, the solution set is $\left\{6\right\}$.