For how many intergers x is the equation `3^(x + 1) = 9^(x -2)` true?

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the equation \( 3^{x + 1} = 9^{x - 2} \) for how many integers \( x \) satisfy it, we can follow these steps:

### Step 1: Rewrite the equation
We know that \( 9 \) can be expressed as \( 3^2 \). Therefore, we can rewrite the equation as:
\[ 3^{x + 1} = (3^2)^{x - 2} \]

### Step 2: Simplify the right-hand side
Using the power of a power property, we simplify the right-hand side:
\[ (3^2)^{x - 2} = 3^{2(x - 2)} = 3^{2x - 4} \]
Now, our equation looks like:
\[ 3^{x + 1} = 3^{2x - 4} \]

### Step 3: Set the exponents equal
Since the bases are the same, we can set the exponents equal to each other:
\[ x + 1 = 2x - 4 \]

### Step 4: Solve for \( x \)
Now, we will solve for \( x \):
1. Rearrange the equation:
\[ 1 + 4 = 2x - x \]
This simplifies to:
\[ 5 = x \]

### Step 5: Conclusion
Thus, the only integer solution for \( x \) is:
\[ x = 5 \]

### Step 6: Count the integer solutions
Since we found only one integer solution, we conclude that there is exactly **1 integer** \( x \) that satisfies the equation.

### Final Answer
The number of integers \( x \) for which the equation \( 3^{x + 1} = 9^{x - 2} \) is true is **1**.

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