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

1 Answer
May 29, 2016

Answer:

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

Explanation:

#(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!