How do you solve $log_x 25=-0.5$ ?

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Answer 1 · 2015-12-01T07:39:05

$x = 1/625$

Using the property $x^(log_x(a)) = a$ we get

$log_x(25) = -0.5$

$=> x^(log_x(25)) = x^(-0.5)$

$=> 25 = 1/sqrt(x)$

$=> sqrt(x) = 1/25$

$=> x = (1/25)^2 = 1/625$

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