How do you solve $4 log x = log (\frac{1}{16})$ $4logx= log (1/16)$ ?

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Answer 1 · 2016-03-17T10:41:00

$x = \frac{1}{2}$ $x=1/2$

As $4 log x = log (\frac{1}{16})$ $4logx=log(1/16)$

${log x}^{4} = log (\frac{1}{2^{4}})$ $logx^4=log(1/2^4)$ or

$x^{4} = 2^{- 1 \times 4} = {(2^{- 1})}^{4}$ $x^4=2^(-1xx4)=(2^(-1))^4$

Hence $x = 2^{- 1}$ $x=2^(-1)$ or $x = \frac{1}{2}$ $x=1/2$

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