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How do you solve for #log_4 x = -2#?

1 Answer

Shwetank Mauria

Jul 21, 2018

#x=1/16#

Explanation:

From the definition of logarithm, if #log_ab=c# then #b=a^c#

Hence as #log_4x=-2#,

we have #4^(-2)=x#

i.e #x=4^(-2)=1/4^2=1/16#

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