How do you solve $(1/2) ^(3x)=8 ^2$ ?

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Answer 1 · 2016-05-06T01:03:36

I found $x=-2$

We can try by taking $log_2$ on both sides and applying some properties:

$log_2(1/2)^(3x)=log_2(8^2)$

$color(red)(3x)[log_2(1)color(blue)(-)log_2(2)]=log_2(64)$

and using the definition of log:

$3x[0-1]=6$

$-3x=6$

$x=-6/3=-2$

Answer 2 · 2016-05-06T17:53:12

Treat as an exponential equation.

$x = -2$

This question is probably easier than it would appear at first. The clue is that 8 is a power of 2.

$(1/2)^(3x) = 8^2$

Make the bases the same.

$(2^(-1))^(3x) = (2^3)^2$

$2^(-3x) = 2^6$

If the bases are equal then the indices are equal.

$-3x = 6$ $rArr x = -2$

How do you solve (1/2) ^(3x)=8 ^2(1/2) ^(3x)=8 ^2?