Fractional Exponents and Radicals: Practice

Take the fourth root (denominator $4$), then cube (numerator $3$).

$16^{3/4}=\left(\sqrt[4]{16}\right)^3=2^3=8$

Pull out the largest perfect-square factor of $50$, which is $25$, since $50=25\cdot 2$.

$\sqrt{50}=\sqrt{25}\,\sqrt{2}=5\sqrt{2}$

The index $3$ becomes the denominator and the inside power $2$ becomes the numerator.

$\sqrt[3]{x^2}=x^{2/3}$

These are like radicals, so add their coefficients, keeping the shared $\sqrt{5}$.

$2\sqrt{5}+3\sqrt{5}=(2+3)\sqrt{5}=5\sqrt{5}$

The radicand does not change, so it is not $5\sqrt{10}$.

The minus sign asks for a reciprocal, and the $\tfrac{1}{2}$ power is a square root.

$25^{-1/2}=\frac{1}{\sqrt{25}}=\frac{1}{5}$

The value is a positive fraction, not a negative number.

The largest perfect-square factor of $12$ is $4$, since $12=4\cdot 3$.

$\sqrt{12}=\sqrt{4}\,\sqrt{3}=2\sqrt{3}$

A root of index $4$ is a one-over-four power.

$\sqrt[4]{a}=a^{1/4}$

Simplify $\sqrt{8}$ first, then combine like radicals. Since $8=4\cdot 2$, $\sqrt{8}=2\sqrt{2}$.

$\sqrt{8}+\sqrt{2}=2\sqrt{2}+\sqrt{2}=3\sqrt{2}$

Roots do not split across a sum, so it is not $\sqrt{10}$.

The largest perfect-square factor of $18$ is $9$, since $18=9\cdot 2$.

$\sqrt{18}=\sqrt{9}\,\sqrt{2}=3\sqrt{2}$

Take the square root (denominator $2$), then cube (numerator $3$).

$4^{3/2}=\left(\sqrt{4}\right)^3=2^3=8$

The index $5$ becomes the denominator and the inside power $3$ becomes the numerator.

$\sqrt[5]{7^3}=7^{3/5}$

Take the cube root (denominator $3$), then square (numerator $2$).

$27^{2/3}=\left(\sqrt[3]{27}\right)^2=3^2=9$