Question

PLEASE HELP ME IN THIS QUESTION!!!!!​

Answer 1

${2}^{ - m} \times \frac{1}{ {2}^{m} } = \frac{1}{4} \\ {2}^{ - m} \times {2}^{ - m} = {2}^{ - 2} \\ \\ {2}^{ - 2m} = {2}^{ - 2} \\ - 2m = - 2 \\ m = 1$
$\frac{1}{14}( ( { {4}^{m} )}^{ \frac{1}{2} } + {( \frac{1}{ {5}^{m} } )}^{ - 1} ) \\ \frac{1}{14} ( {4}^{ \frac{1}{2} } + 5) \\ \frac{1}{14} (2 + 5) \\ \frac{1}{14} \times 7 \\ \frac{1}{2}$

Answer 2

Answer:

$2 {}^{ - m} \times \frac{1}{2 {}^{m} } = \frac{1}{4} \\ \frac{1}{2 {}^{m} } \times \frac{1}{2 {}^{m} } = \frac{1}{4} \\ ie \: \: \: \: ( \frac{1}{2 {}^{m} } ) {}^{2} = \frac{1}{4} \\ ie \: \: \: \: \frac{1}{2 {}^{m} } = \sqrt \frac{1}{4} = \frac{1}{2 } \\ here \: base \: are \: same \: thus \: power \: is \: equal \\ means \: m = 1 \\ hence \: as \: per \: question \\ \frac{1}{14 } \times ((4 {}^{m} ){}^{ \frac{1}{2} } + ( \frac{1}{5 {}^{m} } ) {}^{ - 1} ) \\ substitute \: here \: m = 1 \: we \: will \: get \\ \frac{1}{14} \times (2 + 5) = \frac{1}{2} \\ thus \:the \: required \: answer \: is \: \frac{1}{2} .$