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Question : 51 of 86

Marks: +1, -0

If

\; \frac{9^{y} \cdot 3^{2}(3^{-y})^{-1} - 27^{y}}{3^{3x} \cdot 2^{3}} = \frac{1}{27}

x

-y is given by;

0
$-1$
3
1 .

Solution:

Concept:

Use laws of indices to simplify the equation and solve for

x - y

.

Explanation:

Simplify the numerator:

9^{y} = 3^{2y}

,

3^{2} = 3^{2}

,

(3^{-y})^{-1} = 3^{y}

.

Product:

3^{2y} \cdot 3^{2} \cdot 3^{y} = 3^{2y+2+y} = 3^{3y+2}

.

27^{y} = (3^{3})^{y} = 3^{3y}

.

So numerator =

3^{3y+2} - 3^{3y} = 3^{3y}(3^{2} - 1) = 3^{3y} \cdot 8

.

Denominator:

3^{3x} \cdot 2^{3} = 3^{3x} \cdot 8

.

Left side becomes

\frac{8 \cdot 3^{3y}}{8 \cdot 3^{3x}} = 3^{3y-3x} = 3^{3(y-x)}

.

Equation:

3^{3(y-x)} = \frac{1}{27} = 3^{-3}

.

Thus

3(y-x) = -3 \implies y-x = -1 \implies x-y = 1

.

Answer:

x - y = 1

.

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