https://file.aiquickdraw.com/mathbot/file/17270035491496973ujfp.jpg

The equation shown in your image is:

\[
(x - 3)^{x + 9} = (x - 3)^{2x + 8}
\]

To solve this equation, we can observe that the bases on both sides are the same, i.e., \( (x - 3) \). Therefore, we can equate the exponents:

\[
x + 9 = 2x + 8
\]

Now, solving for \( x \):

1. Subtract \( x \) from both sides:
   \[
   9 = x + 8
   \]

2. Subtract 8 from both sides:
   \[
   1 = x
   \]

Thus, the solution is \( x = 1 \).

However, we must check if \( x = 1 \) is valid for the original equation. When we substitute \( x = 1 \) into \( (x - 3) \), we get:

\[
(1 - 3) = -2
\]

Since the base of an exponent cannot be negative for real numbers, \( x = 1 \) is **not a valid solution** in this case.

Therefore, there is no solution to the equation where \( x - 3 \) is real and positive.

Would you like more details or have any further questions?

Here are 5 related questions:
1. How can you solve exponential equations with different bases?
2. What happens when the base of an exponential equation is negative?
3. Are there any solutions if complex numbers are allowed?
4. Can logarithms be used to solve this kind of equation?
5. What types of equations are best solved by equating exponents?

**Tip:** Always check the domain of your solution in exponential equations, as some solutions may result in invalid expressions.

Learn how to solve an exponential equation where the bases are the same: (x - 3)^(x + 9) = (x - 3)^(2x + 8). This solution involves equating exponents, solving for x, and checking the domain for valid solutions. Suitable for high school students in Grades 10-12.

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