How do you solve #Log_(1/4) (x) + log _2 (x^2) =3#?

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Answer 1 · 2016-05-29T01:08:25

We must first put in a common base; this can be done by using the change of base rule #log_a(n) = logn/loga#

#(logx)/log(1/4) + logx^2/log2 = 3#

#logx/log(2^-2) + logx^2/log2 = 3#

#logx/(-2log2) + logx^2/log2 = 3#

#logx/(-2log2) + (-2logx^2)/(-2log2) = 3#

#log_(1/4)(x) + log_(1/4)(x^-4) = 3#

#log_(1/4)(x xx x^-4) = 3#

#x^-3 = (1/4)^3#

#1/x^3 = 1/64#

#x^3 = 64#

#x = 4#

Hopefully this helps!