# How do you solve x^(4/3)+9=25?

Jan 9, 2016

$x = 8$

#### Explanation:

Firstly, we rearrange the equation so that there will only be equation with one unknown on one side while integer on the other side.

So,
${x}^{\frac{4}{3}} + 9 = 25$

${x}^{\frac{4}{3}} = 25 - 9$

${x}^{\frac{4}{3}} = 16$

To eliminate the power on unknown $x$, we need to whether add to the power of or to the power root of on both side of the equation. For example;

${x}^{\frac{4}{3}} = 16$ also equals to ${\sqrt[3]{x}}^{4} = 16$

Let say we want to eliminate the power root of $3$ on $x$, we must add to the power of $3$ onto both side of equation;

${x}^{\frac{4}{3}} = 16$
Add to the power of $3$ on both side of equation;

${x}^{\left(\frac{4}{3}\right) \left(3\right)} = {16}^{3}$

${x}^{4} = 4096$

Let say then we want to eliminate the power of $4$ on $x$, we must add to the power root of $4$ onto both side of equation;

${x}^{4} = 4096$
Add to the power root of $4$ on both side of equation;

${x}^{\left(4\right) \left(\frac{1}{4}\right)} = {4096}^{\frac{1}{4}}$

$x = 8$