Question

$[( x^{a- a^{-1}} )^{ \frac{1}{a-1} }}] ^{ \frac{a}{a+1} } = x$

Answer 1

Solution:

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Given:

$[(x^{a - a^{-1}}) ^{ \frac{1}{a-1} }]^ \frac{a}{a+1} = x$

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To Prove:

LHS = RHS

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As we know the power rules,

${x^{y}}^{z} = x^{y^{z}}$,

We substitute here,



=> $[(x^{a - a^{-1}}) ^{ \frac{1}{a-1} }]^ \frac{a}{a+1} = x$

=> $[(x^{a - \frac{1}{a} }) ^{ \frac{1}{a-1} }]^ \frac{a}{a+1} = x$

=> $[(x^{ \frac{a^2-1}{a} }) ^{ \frac{1}{a-1} }]^ \frac{a}{a+1} = x$

=> $x^{ \frac{(a+1)(a-1)}{a} } ^{( \frac{1}{a-1}) }^{(\frac{a}{a+1})} = x$

=> $x^{ \frac{a}{a} } = x$

=> $x^1 = x$

=> x = x,

=> LHS = RHS,

Hence, proved.
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                                  Hope it Helps!!

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Answer 2

I'm really sorry I couldn't give u the right answer