Rational Exponents: Practice

The definition says the denominator of the exponent is the index of the root, so a square root is the power $\tfrac12$.

$\sqrt{7} = 7^{1/2}$

The check is the power law: $\left(7^{1/2}\right)^{2} = 7^{1} = 7$, which is exactly what it means to be a square root of $7$.

The exponent $\tfrac12$ asks for the principal square root, the nonnegative number whose square is $9$.

$9^{1/2} = \sqrt{9} = 3$

The exponent is not a divisor: $9 \div 2 = 4.5$ is the trap.

The exponent $\tfrac13$ asks for the cube root, the number whose third power is $27$.

$27^{1/3} = \sqrt[3]{27} = 3 \quad \text{because} \quad 3^{3} = 27$

The denominator $3$ is the root and the numerator $2$ is the power. Take the root first to keep the numbers small.

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

The other route agrees: $\sqrt[3]{8^{2}} = \sqrt[3]{64} = 4$.

The denominator of the exponent is the index of the root and the numerator is the power.

$x^{3/4} = \sqrt[4]{x^{3}}$

The others are different powers: $\sqrt[3]{x^{4}} = x^{4/3}$, $\left(\sqrt{x}\right)^{3} = x^{3/2}$, and $\sqrt{x^{3/4}} = x^{3/8}$.

The denominator $4$ calls for a fourth root, and $\sqrt[4]{16} = 2$ because $2^{4} = 16$.

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

The trap is $16 \cdot \tfrac34 = 12$, which treats the exponent as a multiplier.

Take the cube root first, then square.

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

The power-first route lands in the same place, at the cost of bigger numbers: $\sqrt[3]{27^{2}} = \sqrt[3]{729} = 9$.

The exponent $\tfrac15$ asks for the fifth root of $32$.

$32^{1/5} = \sqrt[5]{32} = 2 \quad \text{because} \quad 2^{5} = 32$

Dividing instead of rooting would give $32 \div 5 = 6.4$, which is the mistake to avoid.

The denominator $2$ is the index of the root and the numerator $5$ is the power, so take the square root and raise it to the fifth.

$a^{5/2} = \left(a^{1/2}\right)^{5} = \left(\sqrt{a}\right)^{5}$

The others are different powers: $\sqrt[5]{a^{2}} = a^{2/5}$ and $\sqrt{a^{5/2}} = a^{5/4}$.

A rational power of a fraction acts on the numerator and the denominator separately, since both are positive.

$\left(\frac{1}{9}\right)^{1/2} = \frac{1^{1/2}}{9^{1/2}} = \frac{1}{3}$

The square root of a number between $0$ and $1$ is larger than the number itself, but still less than $1$.

A negative exponent means take the reciprocal. Deal with the sign first, then the root.

$8^{-1/3} = \frac{1}{8^{1/3}} = \frac{1}{\sqrt[3]{8}} = \frac{1}{2}$

The base $8$ is positive, so the answer is positive. A negative exponent flips the value over; it never changes its sign.

Multiplying powers of the same base adds the exponents, so add the two fractions over a common denominator of $6$.

$x^{1/2} \cdot x^{1/3} = x^{\frac12 + \frac13} = x^{\frac36 + \frac26} = x^{5/6}$

Multiplying the exponents would give $x^{1/6}$, but that is what raising a power to a power does, not what multiplying two powers does.