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

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Answer 1 · 2018-07-25T18:52:25

#x=0#

Given that

#(1/4)^{2x}=(1/2)^x#

#(1/2^2)^{2x}=(1/2)^x#

#(1/2)^{2\cdot 2x}=(1/2)^x#

#(1/2)^{4x}=(1/2)^x#

Comparing the powers of base #1/2# on both the sides we get

#4x=2x#

#2x=0#

#x=0#

Answer 2 · 2018-07-25T19:34:27

#x=0#

#(1/4)^(2x)=((1/2)^2)^(2x)=(1/2)^(4x)#

#"For "(1/2)^(4x)=(1/2)^xrArrx=0#