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We are given the equation \(6^x = 9\). We need to calculate the following:

### (a) \( 6^{-x} \)

We know that \( 6^x = 9 \), so \( 6^{-x} \) is the reciprocal of \( 6^x \):

\[
6^{-x} = \frac{1}{6^x} = \frac{1}{9}.
\]

Thus, the answer is \( \frac{1}{9} \).

### (b) \( \frac{x}{6^2} \)

First, calculate \( 6^2 \):

\[
6^2 = 36.
\]

Now, since \( x \) is unknown, we express \( \frac{x}{6^2} \) as:

\[
\frac{x}{6^2} = \frac{x}{36}.
\]

Given that \( x = \log_6 9 \), we can approximate \( x \) using logarithms, but without a calculator, we leave it as \( \frac{x}{36} \).

### (c) \( 6^0 + 6^x \)

We know:

\[
6^0 = 1 \quad \text{and} \quad 6^x = 9.
\]

So,

\[
6^0 + 6^x = 1 + 9 = 10.
\]

Thus, the answer is 10.

Would you like any further clarification on any part of the solution?

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### Related Questions:
1. How do you simplify expressions with negative exponents?
2. What is the reciprocal of a number and how is it calculated?
3. How can logarithms help in solving equations with exponents?
4. What is the value of \(6^x\) when \(x = 2\)?
5. How do you simplify an expression like \(6^0 + 6^x\) using exponent rules?

### Tip:
For any base \(a\), the expression \(a^0 = 1\), regardless of the value of \(a\) (as long as \(a \neq 0\)).

Given that 6^x = 9, write down the value of (a) 6^(-x), (b) x/6^2, and (c) 6^0 + 6^x.

Learn how to solve equations involving exponents using 6^x = 9. This includes calculating 6^(-x), x/6^2, and 6^0 + 6^x, applying exponent rules and reciprocal properties. Suitable for students in Grades 8-10.

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