To solve the equation $4^{x-7} = 9^{-2x}$, we can start by rewriting both sides with a common base if possible, or use logarithms.

1. Rewriting bases: We notice $4$ and $9$ are powers of $2$ and $3$ respectively, but since these do not match and $9$ is not a power of $4$, we will use logarithms.
2. Applying logarithms: $\log(4^{x-7}) = \log(9^{-2x})$
3. Using the power rule of logarithms: $(x-7) \log(4) = -2x \log(9)$
4. Substitute $\log(4)$ and $\log(9)$ using $\log(4) = 2\log(2)$ and $\log(9) = 2\log(3)$: $(x-7) \cdot 2\log(2) = -2x \cdot 2\log(3)$
5. Simplify: $(x-7)\log(2) = -x\log(3)$
6. Expand and solve for $x$: $x \log(2) - 7 \log(2) = -x \log(3)$ $x \log(2) + x \log(3) = 7 \log(2)$ $x(\log(2) + \log(3)) = 7 \log(2)$ $x = \frac{7 \log(2)}{\log(2) + \log(3)}$

Let's calculate the value of $x$ using the above formula.The solution to the equation $4^{x-7} = 9^{-2x}$ is $x \approx 2.708$ when rounded to the nearest thousandth.

Would you like to know more details about any step of the solution or have any other questions? Here are some topics you might consider:

1. How to apply logarithms to both sides of an equation?
2. Why is it necessary to use logarithms for this problem?
3. What are the logarithmic properties used in solving this equation?
4. How would the solution change if the equation constants were different?
5. What is the significance of choosing a particular base for logarithms in this context?

Tip: When solving exponential equations where the bases are not the same and cannot be easily converted to the same base, using logarithms can provide a straightforward solution.