How do you solve #0.5^x=16^2#?

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Answer 1 · 2016-09-08T16:02:27

#x = -8#

When you are working with equations with indices, try to

either # rarr " make the bases the same"#

or #rarr " make the indices the same"#

#0.5 = 1/2# This is a better form to use, because you should recognise that 16 is one of powers of 2.......

# rarr2^4 = 16 and 2^8 = 16^2#

#(1/2)^x = 16^2#

#(1/2)^x = 2^8#

#2^-x = 2^8 " " rarr "one of the index laws: "(a/b)^m = (b/a)^-m#

Now the bases are the same so we have:

#-x = 8 hArr x = -8#