Solve for x : `9xx3^(x)=(27)^(2x-5)`

Answer

To solve the equation \( 9 \cdot 3^x = (27)^{2x - 5} \), we can follow these steps:

### Step 1: Rewrite the bases
We know that:
- \( 9 = 3^2 \)
- \( 27 = 3^3 \)

So, we can rewrite the equation as:
\[ 3^2 \cdot 3^x = (3^3)^{2x - 5} \]

### Step 2: Simplify both sides
Using the property of exponents that states \( a^m \cdot a^n = a^{m+n} \) on the left side, we have:
\[ 3^{2 + x} = 3^{3(2x - 5)} \]

On the right side, using the property \( (a^m)^n = a^{m \cdot n} \), we get:
\[ 3^{2 + x} = 3^{6x - 15} \]

### Step 3: Set the exponents equal
Since the bases are the same, we can set the exponents equal to each other:
\[ 2 + x = 6x - 15 \]

### Step 4: Solve for \( x \)
Now, we will solve for \( x \):
1. Rearranging the equation:
\[ 2 + 15 = 6x - x \]
This simplifies to:
\[ 17 = 5x \]

2. Dividing both sides by 5:
\[ x = \frac{17}{5} \]

### Final Answer
Thus, the solution for \( x \) is:
\[ x = \frac{17}{5} \]

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