Rational Exponents: Practice

The negative exponent flips the fraction, and then the fourth root acts on the top and bottom separately, since both are positive.

$\left(\frac{16}{81}\right)^{-3/4} = \left(\frac{81}{16}\right)^{3/4} = \left(\sqrt[4]{\frac{81}{16}}\right)^{3} = \left(\frac{3}{2}\right)^{3} = \frac{27}{8}$

Flipping first is what keeps every step positive and small.

Work inside the bracket first, subtracting exponents for each variable.

$x^{-\frac23 - \frac13}\,y^{\frac12 - \left(-\frac12\right)} = x^{-1}y^{1}$

Now raise to the sixth power, multiplying each exponent by $6$.

$\left(x^{-1}y\right)^{6} = x^{-6}y^{6} = \frac{y^{6}}{x^{6}}$

A root inside a root becomes a power of a power, and powers of powers multiply.

$\sqrt[3]{\sqrt{x}} = \left(x^{1/2}\right)^{1/3} = x^{\frac12 \cdot \frac13} = x^{1/6}$

Here the exponents multiply because one root is applied to the other. When you multiply two powers of $x$ the exponents add instead, which would give $x^{5/6}$.

Handle the two factors separately. The factor $\left(\sqrt[3]{2}\right)^{12}$ is a power of a power, so multiply the exponents.

$\left(2^{1/3}\right)^{12} = 2^{4} = 16, \qquad 4^{-1/2} = \frac{1}{\sqrt{4}} = \frac{1}{2}$

Now multiply the results.

$16 \cdot \frac{1}{2} = 8$

Convert both radicals to exponents, then subtract, since dividing powers of one base subtracts the exponents.

$\frac{a^{3/4}}{a^{1/2}} = a^{\frac34 - \frac24} = a^{1/4}$

Dividing the exponents instead would give $\tfrac34 \div \tfrac12 = \tfrac32$, which is the trap.

The proof that the two routes agree ran like this: if $r = \sqrt[n]{a}$, then $r^{m}$ is a positive number whose $n$-th power is $a^{m}$, and a positive number has exactly one positive $n$-th root, so $r^{m} = \sqrt[n]{a^{m}}$. Every step needed $a > 0$.

$\left((-8)^{2}\right)^{1/6} = 2 \quad \text{while} \quad \left((-8)^{1/6}\right)^{2} \text{ is not real}$

That counterexample kills all three of the wider options, since $-8$ is a real number, is nonzero, and is an integer.

Square the binomial as usual, with $u = x^{1/2}$ and $v = x^{-1/2}$, so the expansion is $u^{2} + 2uv + v^{2}$.

The outer terms are $u^{2} = x$ and $v^{2} = x^{-1}$. The cross term is the interesting one, because the exponents cancel.

$2uv = 2 \cdot x^{1/2} \cdot x^{-1/2} = 2x^{0} = 2$

So the expansion is $x + 2 + \dfrac{1}{x}$. Dropping the middle term is the classic error.

For an odd index, a negative number has exactly one real root, and that root is negative.

$(-3)^{3} = -27 \quad \Rightarrow \quad (-27)^{1/3} = \sqrt[3]{-27} = -3$

This is a definition you may use as a name for a real number. What you may not do is push it through the exponent laws, which were proved only for positive bases.

Test each claim. The radical returns the principal root, so $1^{1/2} = 1$ and not $\pm 1$. And $(-1)^{2} = 1$ is correct. And even denominators are perfectly fine for positive bases, as in $16^{1/2} = 4$.

$\left(a^{2}\right)^{1/2} = |a|, \quad \text{which equals } a \text{ only when } a \geq 0$

So the broken link is the middle equality: with $a = -1$ it produces $|-1| = 1$, not $-1$. The rule needs a positive base.

Raising to a positive power preserves order among positive numbers, so raise all four to the $12$th power (the common multiple of the denominators) and compare the results.

$\left(2^{1/2}\right)^{12} = 2^{6} = 64, \quad \left(3^{1/3}\right)^{12} = 3^{4} = 81, \quad \left(8^{1/4}\right)^{12} = 8^{3} = 512, \quad \left(6^{1/6}\right)^{12} = 6^{2} = 36$

The largest twelfth power belongs to the largest number, so $8^{1/4}$ wins.

Work from the inside out. The inner radical is $x^{1/2}$, so the expression under the outer radical is a product of powers of $x$, and their exponents add.

$x \cdot x^{1/2} = x^{3/2}$

Now apply the outer square root, which is the power $\tfrac12$, so the exponents multiply.

$\left(x^{3/2}\right)^{1/2} = x^{3/4}$

Undo the exponent $\tfrac23$ by raising both sides to the reciprocal exponent $\tfrac32$. Both sides are positive, so every law is available.

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

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