How do you solve #4^(-3x)=0.25#?

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Answer 1 · 2016-07-26T21:31:22

#x=1/3#

Since #0.25=1/4=4^-1#

you can write the equivalent equation:

#4^(-3x)=4^-1#

then

#-3x=-1#

#(cancel(-3)x)/cancel(-3)=-1/-3#

#x=1/3#