If `3^(2x - y) = 3^(x+y) = sqrt(27)` , then the value of `3^(x-y) ` will be :

Options

A. `(1)/(sqrt(3))`
B. `(1)/(sqrt(27))`
C. `sqrt(3)`
D. 3

Answer

To solve the problem, we start with the given equations:

1. \( 3^{2x - y} = 3^{x + y} = \sqrt{27} \)

First, we simplify \( \sqrt{27} \):

\[ \sqrt{27} = \sqrt{3^3} = 3^{3/2} \]

Now we can rewrite the equations:

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

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

\[ 2x - y = \frac{3}{2} \quad (1) \]
\[ x + y = \frac{3}{2} \quad (2) \]

Next, we will solve these two equations simultaneously. We can start by solving equation (2) for \( y \):

\[ y = \frac{3}{2} - x \quad (3) \]

Now, substitute equation (3) into equation (1):

\[ 2x - \left(\frac{3}{2} - x\right) = \frac{3}{2} \]

Simplifying this gives:

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

Now, add \( \frac{3}{2} \) to both sides:

\[ 3x = \frac{3}{2} + \frac{3}{2} = 3 \]

Dividing both sides by 3 gives:

\[ x = 1 \]

Now, substitute \( x = 1 \) back into equation (3) to find \( y \):

\[ y = \frac{3}{2} - 1 = \frac{1}{2} \]

Now we have \( x = 1 \) and \( y = \frac{1}{2} \). We need to find \( 3^{x - y} \):

\[ x - y = 1 - \frac{1}{2} = \frac{1}{2} \]

Thus,

\[ 3^{x - y} = 3^{\frac{1}{2}} = \sqrt{3} \]

Therefore, the value of \( 3^{x - y} \) is:

\[ \sqrt{3} \]