Question

solve :
$\frac{2 {}^{x} + 2 {}^{ - x} }{2 {}^{x} - 2 {}^{ - x} } = \frac{16 {}^{x} + 16 {}^{ - x} }{16 {}^{x} - 16 {}^{ - x} }$
plz answer fast
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Answer 1

Answer:

$\sf{The \ value \ of \ x \ is \ 0.}$

Given:

$\sf{\leadsto{\dfrac{2^{x}+2^{-x}}{2^{x}-2^{-x}}=\dfrac{16^{x}+16^{-x}}{16^{x}-16^{-x}}}}$

To find:

$\sf{The \ value \ of \ x.}$

Solution:

$\sf{\leadsto{\dfrac{2^{x}+2^{-x}}{2^{x}-2^{-x}}=\dfrac{16^{x}+16^{-x}}{16^{x}-16^{-x}}}}$

$\sf{By \ componendo}$

$\sf{\leadsto{\dfrac{2(2^{x})}{2^{x}-2^{-x}}=\dfrac{2(16^{x})}{16^{x}-16^{-x}}}}$

$\sf{By \ invertendo}$

$\sf{\leadsto{\dfrac{2^{x}-2^{-x}}{2^{x+1}}=\dfrac{16^{x}-+6^{-x}}{2(16^{x})}}}$

$\sf{But, \ 2^{4}=16 \ substitute \ in \ R.H.S.}$

$\sf{\leadsto{\dfrac{2^{x}-2^{-x}}{2^{x+1}}=\dfrac{2^{4x}-2^{-4x}}{2(2^{4x})}}}$

$\sf{\leadsto{\dfrac{2^{x}-2^{-x}}{2^{x+1}}=\dfrac{2^{4x}-2^{-4x}}{2^{4x+1}}}}$

$\sf{\leadsto{2^{x-x-1}-2^{-x-x-1}=2^{4x-4x-1}-2^{-4x-4x-1}}}$

$\sf{\leadsto{2^{-1}-2^{-2x-1}=2^{-1}-2^{-8x-1}}}$

$\sf{\leadsto{2^{-2x-1}=2^{-8x-1}}}$

$\sf{By \ identity \ If, \ a^{m}=a^{n}, \ then \ m=n}$

$\sf{\leadsto{-2x-1=-8x-1}}$

$\sf{\leadsto{6x=0}}$

$\sf{\leadsto{x=0}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ x \ is \ 0.}}}$