If `9^(x) = sqrt3` and `2^(x+y) = 32` , then y =

Options

A. `(1)/(2)`
B. `(3)/(4)`
C. `(5)/(2)`
D. `(19)/(4)`

Explanation

To solve the equations \( 9^x = \sqrt{3} \) and \( 2^{x+y} = 32 \) for \( y \), we will follow these steps:

### Step 1: Solve for \( x \) from the first equation

Given:
\[ 9^x = \sqrt{3} \]

We can rewrite \( 9 \) as \( 3^2 \):
\[ (3^2)^x = \sqrt{3} \]

Using the property of exponents \( (a^m)^n = a^{m \cdot n} \), we have:
\[ 3^{2x} = \sqrt{3} \]

Now, we can express \( \sqrt{3} \) as \( 3^{1/2} \):
\[ 3^{2x} = 3^{1/2} \]

Since the bases are the same, we can equate the exponents:
\[ 2x = \frac{1}{2} \]

Now, solving for \( x \):
\[ x = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \]

### Step 2: Substitute \( x \) into the second equation

Now we have \( x = \frac{1}{4} \). We will use this value in the second equation:
\[ 2^{x+y} = 32 \]

We know that \( 32 \) can be expressed as \( 2^5 \):
\[ 2^{x+y} = 2^5 \]

Since the bases are the same, we can equate the exponents:
\[ x + y = 5 \]

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

Now we substitute \( x = \frac{1}{4} \) into the equation:
\[ \frac{1}{4} + y = 5 \]

To isolate \( y \), we subtract \( \frac{1}{4} \) from both sides:
\[ y = 5 - \frac{1}{4} \]

To perform the subtraction, we convert \( 5 \) into a fraction with a denominator of \( 4 \):
\[ 5 = \frac{20}{4} \]

Now we can subtract:
\[ y = \frac{20}{4} - \frac{1}{4} = \frac{20 - 1}{4} = \frac{19}{4} \]

### Final Answer
Thus, the value of \( y \) is:
\[ \boxed{\frac{19}{4}} \]