How do you solve $2^{log x} = \frac{1}{4}$ $2^ logx = 1/4$ ?

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How do you solve $2^{log x} = \frac{1}{4}$ $2^ logx = 1/4$ ?

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Answer 1 · 2015-11-27T08:42:02

$x = \frac{1}{100}$ $x = 1/100$ or $x = \frac{1}{e^{2}}$ $x = 1/e^2$ depending on whether $log$ $log$ refers to ${log}_{10}$ $log_10$ or ${log}_{e}$ $log_e$ .

Explanation:

We will use the following:

${log}_{a} (a) = 1$ $log_a(a) = 1$
$log (a^{x}) = x log (a)$ $log(a^x) = xlog(a)$
$a^{{log}_{a} (x)} = x$ $a^(log_a(x)) = x$

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

$\Rightarrow {log}_{2} (2^{log (x)}) = {log}_{2} (2^{- 2})$ $=> log_2(2^log(x)) = log_2(2^(-2))$

$\Rightarrow log (x) {log}_{2} (2) = - 2 {log}_{2} (2)$ $=> log(x)log_2(2) = -2log_2(2)$

$\Rightarrow log (x) = - 2$ $=>log(x) = -2$

(Depending on the context, $log$ $log$ may refer to ${log}_{10}$ $log_10$ or ${log}_{e}$ $log_e$ . We will show both cases.)

$\Rightarrow 10^{{log}_{10} (x)} = 10^{- 2}$ $=> 10^(log_10(x)) = 10^(-2)$ or $e^{{log}_{e} (x)} = e^{- 2}$ $e^(log_e(x)) = e^(-2)$

$\Rightarrow x = 10^{- 2} = \frac{1}{100}$ $=>x = 10^(-2) = 1/100$ or $x = e^{- 2} = \frac{1}{e^{2}}$ $x = e^-2 = 1/e^2$

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Explanation:

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