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How do you solve $8^{2 - x} = 4^{3 x}$ $8^(2-x) = 4^(3x)$ ?

1 Answer

Dec 16, 2015

Express each side of the given equation with a base of $2$ $2$
then equating the exponents, solve for $x$ $x$ to get
$XXX x = \frac{2}{3}$ $color(white)("XXX")x=2/3$

Explanation:

$8^{a} = 2^{3 a}$ $8^a = 2^(3a)$
$4^{b} = 2^{2 b}$ $4^b = 2^(2b)$

Therefore
$XXX 8^{2 - x} = 4^{3 x}$ $color(white)("XXX")8^(2-x) = 4^(3x)$
is equivalent to
$XXX 2^{6 - 3 x} = 4^{6 x}$ $color(white)("XXX")2^(6-3x)=4^(6x)$

Which implies
$XXX 6 - 3 x = 6 x$ $color(white)("XXX")6-3x=6x$

$XXX 9 x = 6$ $color(white)("XXX")9x = 6$

$XXX x = \frac{2}{3}$ $color(white)("XXX")x=2/3$

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