How do you solve #8^{x - 1} = root[3]{16}#?

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Answer 1 · 2016-09-15T19:46:00

#x=13/9#

Note that #8# and #16# are both powers of #2#, so reexpress this in terms of fractional powers of #2#:

#8^(x-1) = (2^3)^(x-1) = 2^(3x-3)#

#root(3)(16) = (2^4)^(1/3) = 2^(4/3)#

So our original equation can be reexpressed as:

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

Concerning ourselves only with Real solutions, the function #f(x) = 2^x# is one to one. So we can deduce:

#3x-3 = 4/3#

Add #3# to both sides to get:

#3x = 13/3#

Divide both sides by #3# to find:

#x = 13/9#