How do you solve for ${log}_{4} x = - 2$ $log_4 x = -2$ ?

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Answer 1 · 2018-07-21T06:35:51

$x = \frac{1}{16}$ $x=1/16$

From the definition of logarithm, if ${log}_{a} b = c$ $log_ab=c$ then $b = a^{c}$ $b=a^c$

Hence as ${log}_{4} x = - 2$ $log_4x=-2$ ,

we have $4^{- 2} = x$ $4^(-2)=x$

i.e $x = 4^{- 2} = \frac{1}{4^{2}} = \frac{1}{16}$ $x=4^(-2)=1/4^2=1/16$

How do you solve for log4x=−2log_4 x = -2?