We are given the logarithmic equation:

$\log_9(x) = -1$

To solve for $x$, we can rewrite this equation in exponential form. The general logarithmic form $\log_b(a) = c$ can be rewritten as $b^c = a$.

Applying this to our equation:

$9^{-1} = x$

Since $9^{-1}$ is the reciprocal of 9, we get:

$x = \frac{1}{9}$

Thus, the solution is:

$x = \frac{1}{9}$

Would you like more details or have any questions?

Here are 5 related questions:

1. How do you convert between logarithmic and exponential forms?
2. What is the logarithmic function and its properties?
3. Can you solve $\log_4(x) = 2$?
4. What is the base-change formula for logarithms?
5. How do you solve logarithmic equations with different bases?

Tip: Always check if the logarithmic argument is valid, ensuring it’s positive.