How do you solve #(1/4)^(2x) = 32^(4x-2)#?

1 Answer
Bdub
Mar 1, 2016

#(2^-2)^(2x)=(2^5)^(4x-2);-> 2^(-4x)=2^(20x-10)#
#-4x=20x-10->24x=10->x=5/12#

Explanation:

To solve for x make sure both sides have the same base. Once they are same then you equate the exponents and solve for x.

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