Find $x$ if $\log_9(2x-7) = \dfrac{3}{2}$.
One way to do that is to use the exponent rule of logarithms, which says that $\log_b(a) = c$ is equivalent to $b^c = a$. Applying that rule to this equation, I get $9^{\dfrac{3}{2}} = 2x-7$. Now I need to simplify the left side. I know that $9$ is the same as $3^2$, so I can rewrite it as $(3^2)^{\dfrac{3}{2}}$. Using the power of a power rule, I can simplify this to $3^3$, which is $27$. So I have $27 = 2x-7$. Adding $7$ to both sides, I get $34 = 2x$. Dividing both sides by $2$, I get $x = 17$.

