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 $x=1 4/9$

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