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.

Discover how to solve the logarithmic equation log9(x) = −1. This problem involves converting between logarithmic and exponential forms. Learn key mathematical concepts, such as the logarithmic formula and its relationship to exponents. Ideal for students in Grades 9-11.

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