How do you solve $log_8 2+log_8(4x^2)=1$ ?

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Answer 1 · 2016-08-04T20:27:25

$x=+-1$ .

$log_8 2+log_8 (4x^2)=1$

$rArr log_8 (2*4x^2)=1$

$rArr 8x^2=8^1=8$

$rArr x^2=1$

$rArr x=+-1$

These roots satisfy the given eqn.

How do you solve log_8 2+log_8(4x^2)=1log_8 2+log_8(4x^2)=1?