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How do you write an equation of a line going through #(1/2)^x =16^2#?

1 Answer

Nov 23, 2016

#x=-8#

Explanation:

#1/2=2^(-1)#

#rArr (1/2)^x=((2)^(-1))^x=2^(-x)#

#16=2^4#

#rArr 16^2 = (2^4)^2=2^8#

Therefore
#color(white)("XXX")(1/2)^x=16^2#

#rArr 2^(-x) = 2^8#

#rArr x=-8#

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