We have the equation:

\[
8^{x-1} = 2^{x+2}
\]

### Step 1: Rewrite the bases as powers of 2

Since \( 8 = 2^3 \), we can rewrite the left-hand side as follows:

\[
8^{x-1} = (2^3)^{x-1}
\]

Now, apply the power of a power rule \( (a^m)^n = a^{m \cdot n} \):

\[
(2^3)^{x-1} = 2^{3(x-1)} = 2^{3x - 3}
\]

Thus, the equation becomes:

\[
2^{3x - 3} = 2^{x + 2}
\]

### Step 2: Set the exponents equal to each other

Since the bases are the same, we can equate the exponents:

\[
3x - 3 = x + 2
\]

### Step 3: Solve for \( x \)

First, subtract \( x \) from both sides:

\[
3x - x - 3 = 2
\]

Simplifying:

\[
2x - 3 = 2
\]

Now, add 3 to both sides:

\[
2x = 5
\]

Finally, divide both sides by 2:

\[
x = \frac{5}{2} = 2.5
\]

### Solution:

The solution is \( x = \frac{5}{2} \) or \( x = 2.5 \).

Would you like further details, or do you have any questions?

Here are five related questions:
1. How would you solve \( 2^{2x+1} = 4^{x+3} \)?
2. What is the general method for solving exponential equations with different bases?
3. How can logarithms be used to solve equations like this?
4. How would you solve \( 8^x = 16^{x-1} \)?
5. What happens if both sides of the equation have irrational exponents?

**Tip**: When solving exponential equations, it’s helpful to express all terms with the same base if possible, which simplifies the comparison of exponents.

This detailed guide explains how to solve the exponential equation 8^{x-1}=2^{x+2} by expressing terms with the same base and equating exponents. Suitable for students in Grades 9-11 studying algebra and exponential functions.

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