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Question : 44 of 86

Marks: +1, -0

On simplification

\left(\,\frac{m^{x}}{m^{y}}\right)^{x+y} \times \left(\,\frac{m^{y}}{m^{Z}}\right)^{y+z}

\div 3 \left(m^{x} m^{2}\right)^{x-2}

reduces to

3
-3
$-\frac{1}{3}$
$\,\frac{1}{3}$

Solution:

Concept:

Use laws of indices to combine powers and simplify the expression to a constant.

Explanation:

First term:

\left(\frac{m^{x}}{m^{y}}\right)^{x+y} = \left(m^{x-y}\right)^{x+y} = m^{(x-y)(x+y)} = m^{x^{2}-y^{2}}

.

Second term:

\left(\frac{m^{y}}{m^{z}}\right)^{y+z} = \left(m^{y-z}\right)^{y+z} = m^{(y-z)(y+z)} = m^{y^{2}-z^{2}}

.

Product of first two:

m^{x^{2}-y^{2}} \cdot m^{y^{2}-z^{2}} = m^{x^{2}-z^{2}}

.

Denominator:

3\left(m^{x}m^{2}\right)^{x-2} = 3\left(m^{x+2}\right)^{x-2} = 3\,m^{(x+2)(x-2)} = 3\,m^{x^{2}-4}

.

Thus the whole expression becomes

\dfrac{m^{x^{2}-z^{2}}}{3\,m^{x^{2}-4}} = \dfrac{1}{3}\,m^{4-z^{2}}

.

For the result to be a constant independent of

x,y,z

, we need

z=2

, giving

\dfrac{1}{3}\,m^{0} = \dfrac{1}{3}

.

Answer:

\dfrac{1}{3}

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