# What is 3 sqrt(x^2/y) in exponential form?

Jul 1, 2016

$\frac{3 x}{y} ^ \left(\frac{1}{2}\right)$

#### Explanation:

$\sqrt{{x}^{2}}$ can be written as $x$, but y does not have a square root.

$3 x {y}^{- \frac{1}{2}}$, but it is better not to have negative exponents.

$\frac{3 x}{y} ^ \left(\frac{1}{2}\right)$

If you wanted to rationalize the denominator:

$\frac{3 x}{y} ^ \left(\frac{1}{2}\right) \times {y}^{\frac{1}{2}} / {y}^{\frac{1}{2}}$

$\frac{3 x {y}^{\frac{1}{2}}}{y}$

Jul 1, 2016

Exponent form: $\pm 3 x {y}^{- \frac{1}{2}}$

#### Explanation:

Here, $y > 0$ for $3 \sqrt{{x}^{2} / y}$ to be real.
The form with exponents is $3 \frac{{\left({x}^{2}\right)}^{\frac{1}{2}}}{y} ^ \left(\frac{1}{2}\right) = 3 \frac{x}{y} ^ \left(\frac{1}{2}\right)$

$= 3 x {y}^{- \frac{1}{2}}$

Jul 3, 2016

$3 \sqrt{{x}^{2} / y} = 3 \left\mid x \right\mid {y}^{- \frac{1}{2}}$ for $y \in \left(0 , \infty\right)$

#### Explanation:

Considering only Real valued square roots, we require:

${x}^{2} / y \ge 0$

Since ${x}^{2} \ge 0$ for any Real $x$, this amounts to $y > 0$.

Note that $\sqrt{{x}^{2}} = \left\mid x \right\mid$ for any Real $x$. The square root sign denotes the principal square root, which in the case of Real square roots is the non-negative one.

Note that if $a \ge 0$ and $b > 0$ then $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$

So we find:

$3 \sqrt{{x}^{2} / y} = 3 \frac{\sqrt{{x}^{2}}}{\sqrt{y}} = 3 \frac{\left\mid x \right\mid}{y} ^ \left(\frac{1}{2}\right) = 3 \left\mid x \right\mid {y}^{- \frac{1}{2}}$