How do you solve # (1/2)^x =16^2#?

1 Answer
Bill K.
Jul 16, 2015

#x=-8#

Explanation:

You could use logarithms, but since #1/2# and #16# are both powers of 2, it's easier if you use that fact.

We have #1/2=2^{-1}# and #16=2^{4}#. Therefore the equation #(1/2)^{x}=16^{2}# can be written as #(2^{-1})^{x}=(2^{4})^{2}#, or #2^{-x}=2^{8}#. This implies that #-x=8# so that #x=-8# (after all, #2^{z}# is a one-to-one function of the variable #z# (just to pick a different letter than #x#)).

You can check the answer like this:

#(1/2)^{-8}=2^{8}=256=16^2#

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