Fractional Exponents and Radicals: Additional Practice (Set 2)

Adding the exponents with the product rule shows $a^{1/2}$ squares to $a$.

$a^{1/2}\cdot a^{1/2}=a^{\frac{1}{2}+\frac{1}{2}}=a^{1}=a$

A number whose square is $a$ is the square root of $a$, so $a^{1/2}=\sqrt{a}$.

The minus sign asks for a reciprocal, and the $\tfrac{1}{4}$ power is a fourth root.

$81^{-1/4}=\frac{1}{\sqrt[4]{81}}=\frac{1}{3}$

Since $3^4=81$, the fourth root is $3$.

The minus sign asks for a reciprocal; then evaluate $8^{2/3}$ as a cube root followed by a square.

$8^{-2/3}=\frac{1}{\left(\sqrt[3]{8}\right)^2}=\frac{1}{2^2}=\frac{1}{4}$

The value is a positive fraction, not $-4$.

Write the fifth root as a one-fifth power and multiply the exponents.

$\left(\sqrt[5]{3}\right)^{10}=\left(3^{1/5}\right)^{10}=3^{10/5}=3^2=9$

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

$100^{3/2}=\left(\sqrt{100}\right)^3=10^3=1000$

Take the square root (denominator $2$), then raise to the fifth (numerator $5$).

$4^{5/2}=\left(\sqrt{4}\right)^5=2^5=32$

Simplify each radical, then subtract the like terms. Here $45=9\cdot 5$ and $20=4\cdot 5$.

$\sqrt{45}-\sqrt{20}=3\sqrt{5}-2\sqrt{5}=\sqrt{5}$

Note $\sqrt{5}$ is not $\sqrt{25}=5$.

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

$27^{4/3}=\left(\sqrt[3]{27}\right)^4=3^4=81$

Pull out the largest perfect-square factor of $200$, which is $100$, since $200=100\cdot 2$.

$\sqrt{200}=\sqrt{100}\,\sqrt{2}=10\sqrt{2}$

Write the fourth root as a one-fourth power and multiply the exponents.

$\left(\sqrt[4]{a}\right)^3=\left(a^{1/4}\right)^3=a^{3/4}$

A quotient of roots is the root of the quotient.

$\frac{\sqrt{72}}{\sqrt{2}}=\sqrt{\frac{72}{2}}=\sqrt{36}=6$

Multiply the radicals, then simplify. Here $\sqrt{2}\cdot\sqrt{6}=\sqrt{12}=2\sqrt{3}$.

$3\sqrt{2}\cdot\sqrt{6}=3\sqrt{12}=3\cdot 2\sqrt{3}=6\sqrt{3}$