If `27^(x)=9^(y-1)`, then

Options

A. `y=(3)/(2)x+1`
B. `y=(3)/(2)x+2`
C. `y=(3)/(2)x+(1)/(2)`
D. `y=(1)/(2)x+(2)/(3)`

Answer

To solve the equation \( 27^x = 9^{y-1} \), we can follow these steps:

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

So we can rewrite the equation as:
\[ (3^3)^x = (3^2)^{y-1} \]

### Step 2: Apply the power of a power property
Using the property \( (a^m)^n = a^{m \cdot n} \), we can simplify both sides:
\[ 3^{3x} = 3^{2(y-1)} \]

### Step 3: Set the exponents equal to each other
Since the bases are the same, we can equate the exponents:
\[ 3x = 2(y - 1) \]

### Step 4: Expand the right side
Distributing the 2 on the right side gives:
\[ 3x = 2y - 2 \]

### Step 5: Rearrange the equation to solve for \( y \)
Add 2 to both sides:
\[ 3x + 2 = 2y \]

Now, divide both sides by 2:
\[ y = \frac{3x + 2}{2} \]

### Step 6: Simplify the expression
We can separate the terms:
\[ y = \frac{3}{2}x + 1 \]

Thus, we have derived that:
\[ y = \frac{3}{2}x + 1 \]

### Conclusion
The correct answer is:
**Option A: \( y = \frac{3}{2}x + 1 \)**

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