How do you solve #4^(x-3)=1/16#?

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Answer 1 · 2016-09-10T07:21:51

#x =1#

There are 2 approaches which I use for exponential equations, which are equations which have indices.

#rarr# make the bases the same
#rarr# make the indices the same

If neither of these work, then use logs.

We should know that 16 is a power of 4, so use 4 as the base on both sides.

#4^(x-3) = 1/16 = 1/4^2#

#4^(x-3) = 4^-2 " "larr# now equate the indices

#x-3 = -2#

#x = -2+3 = 1#