Every question so far handed you the time. Now it is the unknown, and it is stuck up in the exponent where ordinary algebra cannot reach it. A logarithm is the tool built for exactly this: $\log_b x$ asks “what power of $b$ gives $x$?” — so $\log_2 64 = 6$, because six doublings turn 1 into 64.

The move for $a \cdot b^{x} = y$ is two steps. Divide the coefficient away, then take logs of both sides and let the power rule bring the exponent down in front: $x = \dfrac{\ln (y/a)}{\ln b}$ — read aloud, x equals ell-en of y over a, all over ell-en b. It works with $\ln$ or $\log_{10}$; it only matters that BOTH logs are the same kind, because the answer is a ratio and the choice cancels out.

That cancellation is also change of base: $\log_b x = \dfrac{\ln x}{\ln b}$, which is how a calculator with only two log buttons evaluates a logarithm in any base at all. And the trap, named so you can dodge it: $\dfrac{\ln x}{\ln b}$ is NOT $\ln\!\left(\dfrac{x}{b}\right)$. A quotient OF logs and the log OF a quotient are different animals — the second one is a subtraction. Divide the logs, never the numbers.