How do you solve ${log}_{x} 25 = - 0.5$ $log_x 25=-0.5$ ?

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Answer 1 · 2015-12-01T07:39:05

$x = \frac{1}{625}$ $x = 1/625$

Using the property $x^{{log}_{x} (a)} = a$ $x^(log_x(a)) = a$ we get

${log}_{x} (25) = - 0.5$ $log_x(25) = -0.5$

$\Rightarrow x^{{log}_{x} (25)} = x^{- 0.5}$ $=> x^(log_x(25)) = x^(-0.5)$

$\Rightarrow 25 = \frac{1}{\sqrt{x}}$ $=> 25 = 1/sqrt(x)$

$\Rightarrow \sqrt{x} = \frac{1}{25}$ $=> sqrt(x) = 1/25$

$\Rightarrow x = {(\frac{1}{25})}^{2} = \frac{1}{625}$ $=> x = (1/25)^2 = 1/625$

How do you solve logx25=−0.5log_x 25=-0.5?