Solve Logarithm Ratio: Finding log₄ₓ9/log₄ₓa Simplification

Because $\frac{\log_{4x}9}{\log_{4x}a}$ is a ratio of two logarithms, not the logarithm of a ratio! The division is happening outside the logarithm functions, not inside as an argument.

The change-of-base formula is $\log_b x = \frac{\log_a x}{\log_a b}$. You need it here because when you have a ratio of logarithms with the same base, converting to a common base lets the denominators cancel out.

When you apply change-of-base, you get $\frac{\frac{\log_a 9}{\log_a (4x)}}{\frac{\log_a a}{\log_a (4x)}}$. This is the same as $\frac{\log_a 9}{\log_a (4x)} \times \frac{\log_a (4x)}{\log_a a}$, so the $\log_a (4x)$ terms multiply to 1.

Because $\log_a a$ asks "what power of a gives us a?" The answer is always 1, since $a^1 = a$ for any positive number a.

Yes! You can use any valid logarithm base (like base 10 or base e). The key is using the same base for both conversions so the common terms cancel out properly.

Check that your answer makes sense: $\log_a 9$ represents the power you raise a to get 9. This should be independent of the original base $4x$, which our result confirms!