# What is the value of d in the equation root3 (k^d) = (root3 k)^5?

Aug 6, 2018

Given: $\sqrt[3]{{k}^{d}} = {\left(\sqrt[3]{k}\right)}^{5}$

Cube both sides

${k}^{d} = {\left(\sqrt[3]{k}\right)}^{15}$

$\textcolor{b r o w n}{\text{Doing it the 'hard way'}}$

Write ${\left(\sqrt[3]{k}\right)}^{15}$ as

$\left[\sqrt[3]{k} \times \sqrt[3]{k} \times \sqrt[3]{k}\right] \times \left[\sqrt[3]{k} \times \sqrt[3]{k} \times \sqrt[3]{k}\right] \times \left[\sqrt[3]{k} \times \sqrt[3]{k} \times \sqrt[3]{k}\right] \times \left[\sqrt[3]{k} \times \sqrt[3]{k} \times \sqrt[3]{k}\right] \times \left[\sqrt[3]{k} \times \sqrt[3]{k} \times \sqrt[3]{k}\right]$

Which is the same as ${k}^{5}$

Putting it all back together we have:

${k}^{d} = {k}^{5}$

Then by direct comparison we have:

$d = 5$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b r o w n}{\text{Doing it the more straight forward way}}$

Write ${\left(\sqrt[3]{k}\right)}^{15}$ as k^((15/3)

but $15 \div 3 = 5$ giving

${\left(\sqrt[3]{k}\right)}^{15}$ is the same as ${k}^{5}$

Then the rest is as above.

Aug 7, 2018

color(orange)(d = 5

#### Explanation:

${\sqrt[3]{k}}^{d} = {\left(\sqrt[3]{k}\right)}^{5}$

${k}^{\frac{d}{3}} = {\left({k}^{\frac{1}{3}}\right)}^{5}$

${k}^{\frac{d}{3}} = {k}^{\frac{5}{3}}$

$\frac{d}{\cancel{3}} = \frac{5}{\cancel{3}}$

color(orange)(d = 5