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The equation provided is:

\[
7^{x - 3} = 7^{3x - 6}
\]

To solve for \( x \), we can equate the exponents, as the bases on both sides of the equation are the same:

\[
x - 3 = 3x - 6
\]

Now, solve for \( x \):

1. Move \( x \) terms to one side by subtracting \( x \) from both sides:
   \[
   -3 = 2x - 6
   \]

2. Add 6 to both sides:
   \[
   3 = 2x
   \]

3. Divide by 2:
   \[
   x = \frac{3}{2}
   \]

So, the solution is:
\[
x = \frac{3}{2}
\]

Would you like further details or have any questions?

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Here are some related questions for practice:

1. How would you solve \( 5^{2x + 1} = 5^{4x - 3} \)?
2. What happens if the bases are different in an equation like \( 3^{x + 1} = 4^{2x - 5} \)?
3. How can logarithms be used to solve exponential equations with different bases?
4. How would you check if \( x = \frac{3}{2} \) is correct in this equation?
5. Can you solve \( 2^{x + 3} = 16 \) for \( x \) using a similar method?

**Tip:** When the bases of exponents are the same, you can simplify the equation by equating the exponents directly.

Solve for x in the equation 7^(x - 3) = 7^(3x - 6).

Learn how to solve the exponential equation 7^(x - 3) = 7^(3x - 6) by equating exponents, a key algebra concept. This step-by-step solution is ideal for students in Grades 7-8.

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