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.