How do you solve z^(-1/4) = 1/3?

Jan 18, 2017

${z}^{- \frac{1}{4}} = \frac{1}{3} \iff z = 81$

Explanation:

${z}^{- \frac{1}{4}} = \frac{1}{3}$

$\iff$ First since ${x}^{- 1} = \frac{1}{x}$, and ${x}^{a b} = {\left({x}^{a}\right)}^{b}$

${z}^{\left(- 1\right) \left(\frac{1}{4}\right)} = {\left({z}^{\frac{1}{4}}\right)}^{- 1} = \frac{1}{z} ^ \left(\frac{1}{4}\right) = \frac{1}{3}$

$\iff$

$1 = {z}^{\frac{1}{4}} / 3$ multiply both sides by ${z}^{\frac{1}{4}}$

$\iff$

$3 = {z}^{\frac{1}{4}}$ multiply both sides by 3

$\iff$

${3}^{4} = {\left({z}^{\frac{1}{4}}\right)}^{4}$ raise both sides to the power of 4

$\iff$

${3}^{4} = z$

$\iff$

${\left({3}^{2}\right)}^{2} = \left({3}^{2}\right) \left({3}^{2}\right) = 9 \left(9\right) = \underline{81 = z}$