# How do you solve 2x^(3/4)=54?

Jul 31, 2016

$x = 81$

#### Explanation:

Divide both sides by $2$ to get:

${x}^{\frac{3}{4}} = 27 = {3}^{3}$

If $x$ is Real and positive then so is ${x}^{\frac{3}{4}}$ and we can raise both sides to the power $\frac{4}{3}$ to find:

$x = {x}^{1} = {x}^{\frac{3}{4} \cdot \frac{4}{3}} = {\left({x}^{\frac{3}{4}}\right)}^{\frac{4}{3}} = {\left({3}^{3}\right)}^{\frac{4}{3}} = {3}^{3 \cdot \frac{4}{3}} = {3}^{4} = 81$

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Other solutions?

Are there any negative or Complex solutions?

Suppose $x = r \left(\cos \theta + i \sin \theta\right)$ where $r > 0$ and $\theta \in \left(- \pi , \pi\right]$

Then:

${x}^{\frac{3}{4}} = {r}^{\frac{3}{4}} \left(\cos \left(\frac{3 \theta}{4}\right) + i \sin \left(\frac{3 \theta}{4}\right)\right)$

For the imaginary part to be $0$, we must have $\sin \left(\frac{3 \theta}{4}\right) = 0$

So $\theta = \frac{4 \pi}{3} k$ for some integer $k$

This can only lie in the range $\left(- \pi , \pi\right]$ if $k = 0$, so $\theta = 0$.

So the only solution is the one on the positive part of the Real axis, i.e. $81$.