Question

if 3^x/1+3^x =1/3 then value of 27^x/1-27^x is

Answer 1

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$= > \frac{ {3}^{x} }{1 + {3}^{x} } = \frac{1}{3} \\ \\ = > \frac{1 + {3}^{x} }{ {3}^{x} } = 3 \\ \\ = > \frac{1}{ {3}^{x} } + 1 = 3 \\ \\ = > \frac{1}{ {3}^{x} } = 3 - 1 \\ \\ = > \frac{1}{ {3}^{x} } = 2 \\ \\ = > \boxed{ \bf {3}^{x} = \frac{1}{2} } \\ \\ \: \: \: \: \: \: \bf \underline{cubing \: \: both \: \: sides} \\ \\ = > {( {3}^{x} )}^{3} = {( \frac{1}{2} )}^{3} \\ \\ = > {3}^{3x} = \frac{1}{8} \\ \\ = > \boxed{ \bf{27}^{x} = \frac{1}{8} } \\ \\ = = = = = = = = = = = = = = = \\ \: \: \: \: \: \: \: \underline{now} \: \\ \\ = > \frac{ {27}^{x} }{1 - {27}^{x} } \\ \\ = \frac{ \frac{1}{8} }{1 - \frac{1}{8} } = \frac{ \frac{1}{8} }{ \frac{7}{8} } \\ \\ = \frac{1}{8} \times \frac{8}{7} = \boxed{ \bf \: \underline{\frac{1}{7}} }$
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Answer 2

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$\underline{SOLUTION}$ ➡

$\frac{3 {}^{x} }{1 + 3 {}^{x} } = \frac{1}{3} \\ \\ = > 3 {}^{x} \times 3 = 1 + 3 {}^{x} \\ \\ = > 3 = \frac{1 + 3 {}^{x} }{3 {}^{x} } \\ \\ = > \frac{1 + 3 {}^{x} }{3 {}^{x} } = 3 \\ \\ = > \frac{1}{3 {}^{x} } + \frac{3 {}^{x} }{3 {}^{x} } = 3 \\ \\ = > \frac{1}{3 {}^{x} } + 1 = 3 \\ \\ = > \frac{1}{3 {}^{x} } = 3 - 1 \\ \\ = > \frac{1}{3 {}^{x} } = 2 \\ \\ = > 3 {}^{x} = \frac{1}{2} \: ----->(1)$

$By \: \: cubing \: \: both \: \: sides \: \: of (1) ,$

$(3 {}^{x} ) {}^{3} = ( \frac{1}{2} ) {}^{3} \\ \\ = >( 3 {}^{3} ) {}^{x} = \frac{1}{8} \\ \\ = > 27 {}^{x} = \frac{1}{8}$

$Now, \\ \\ By \: \: putting \: \: the \: \: value \: \: of \: \: \boxed{27 {}^{x} = \frac{1}{8} }\: \: \\ in \: \: \frac{27 {}^{x} }{1 - 27 {}^{x} }, \: \: we \: \: get ,\\ \\ \\ \frac{27 {}^{x} }{1 - 27 {}^{x} } \\ \\ = \frac{ \frac{1}{8} }{1 - \frac{1}{8} } \\ \\ = \frac{ \frac{1}{8} }{ \frac{8 - 1}{8} } \\ \\ = \frac{ \frac{1}{8} }{ \frac{7}{8} } \\ \\ = \frac{1}{8} \times \frac{8}{7} \\ \\ = \frac{1}{7}$

$\underline{ANSWER}$ ➡

$\boxed{\huge{ \frac{27 {}^{x} }{1 - 27 {}^{x} } = \frac{1}{7} }}$

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