How do you solve #1/32^(2x)=64#?

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Answer 1 · 2016-04-25T15:52:46

#x=-3/5#

Using properties of exponents we are going to rewrite everything in terms of base #2#.

#1/(32^2)^x=64#

#1/((2^5)^2)^x=64#

#1/2^(10x)=2^6#

#2^(-10x)=2^6#

Since we are now in the same base we can do the following

#-10x=6#

#x=6/-10=-3/5#