How do you solve $x^{- 3} = \frac{1}{8}$ $x^-3 = 1/8$ ?

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Answer 1 · 2016-03-16T03:27:26

$x = 2$ $x=2$

$x^{- 3} = \frac{1}{8}$ $color(blue)(x^-3=1/8$

Use the law of Negative exponents $x^{- y} = \frac{1}{x^{y}}$ $color(brown)(x^-y=1/x^y$
$:.x^-3=1/x^3$

$rarr1/x^3=1/8$

Cross multiply $color(orange)(a/b=c/d,ad=bc$

$rarrx^3=8$

$rArrcolor(green)(x=root(3)8=2$

How do you solve x^-3 = 1/8x−3=18x^-3 = 1/8?