Solve the Inverse Logarithm: Finding (log₇x)⁻¹

The inverse logarithm property comes from the definition of logarithms. If $\log_7 x = y$, then $7^y = x$. The inverse asks: what power do we raise x to get 7? That's exactly $\log_x 7$!

No! The notation $(\log_7 x)^{-1}$ means the inverse function, not the reciprocal. If you wanted the reciprocal, you'd write $\frac{1}{\log_7 x}$, which is completely different.

Think of it as a "position swap": whatever is the base becomes the argument, and whatever is the argument becomes the base. Base and argument trade places!

Yes! The property $(\log_b a)^{-1} = \log_a b$ works for any valid base (positive, not equal to 1) and any positive argument.

Yes! Use the change of base formula: $\log_x 7 = \frac{\ln 7}{\ln x}$. Multiply this by $\log_7 x = \frac{\ln x}{\ln 7}$ and you should get 1.