How do you solve #(1/4)^(2x)=(1/32)^(3x+1)#?

1 Answer
Cesareo R.
May 26, 2016

#x = -5/11#

Explanation:

#(1/4)^{2x}=(1/32)^{3x+1} equiv 1/(4)^{2x}=1/(32)^{3x+1}#
# 1/(4)^{2x}=1/(32)^{3x+1} equiv (4)^{2x}=(32)^{3x+1} #
# (4)^{2x}=(32)^{3x+1} equiv (2^2)^{2x}=(2^5)^{3x+1}#
#(2^2)^{2x}=(2^5)^{3x+1}equiv 2^{2 times 2 x}=2^{5 times(3x+1)}#
# 2^{2 times 2 x}=2^{5 times(3x+1)}equiv 4x=15x+5#
solving for #x# we get #x = -5/11#

Related questions
See all questions in Logarithm-- Inverse of an Exponential Function
Impact of this question
1745 views around the world
You can reuse this answer
Creative Commons License