How do you solve #8^(x-1)=root3(16)#?

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Answer 1 · 2016-05-24T17:30:32

#x=13/9# or #x=1 4/9#

To solve #8^(x-1)=root(3)16#

We begin by taking the #16# out of the root

#8^(x-1)=16^(1/3)#

Next convert the #8# and #16# to the same base.

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

Now simplify the exponents

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

Since the base values are the same we can simply simplify the exponents algebraically.

#3x-3 =4/3#

#3xcancel(-3) cancel(+3) =4/3 + 3#

#3x = 13/3#

#cancel3x (1/cancel3)= 13/3(1/3)#

#x=13/9# or #x=1 4/9#