Rational Exponents: Additional Practice (Set 2)

A single expression cannot name two different numbers, so something in the chain is not licensed. Check the endpoints: $(-8)^{1/3} = -2$ is a correct use of the odd-root convention, and $64^{1/6} = 2$ is correct, and $\tfrac26$ and $\tfrac13$ are the same rational number.

$\left((-8)^{2}\right)^{1/6} = 2 \quad \text{but} \quad (-8)^{1/3} = -2$

What fails is the step that treats both as $(-8)^{2/6}$. Well-definedness was proved from the fact that a positive number has exactly one positive $n$-th root, and $-8$ has none. The hypothesis $a > 0$ is exactly what the proof consumed.

The factor is the ratio of the two edge lengths, and both volumes are positive, so the exponent distributes over the quotient.

$\frac{128^{1/3}}{64^{1/3}} = \left(\frac{128}{64}\right)^{1/3} = 2^{1/3}$

Doubling the volume multiplies the edge by only the cube root of $2$, roughly $1.26$. The edge grows far more slowly than the volume, which is why the answer is not $2$.

Take the two terms one at a time. The negative exponent flips the fraction before the square root acts.

$\left(\frac{25}{4}\right)^{-1/2} = \left(\frac{4}{25}\right)^{1/2} = \frac{2}{5}, \qquad 16^{3/4} = \left(\sqrt[4]{16}\right)^{3} = 8$

Now add, writing $8$ as $\tfrac{40}{5}$.

$\frac{2}{5} + \frac{40}{5} = \frac{42}{5}$

Undo the exponent $\tfrac32$ by raising both sides to its reciprocal $\tfrac23$, which is legal because both sides are positive.

$b = \left(b^{3/2}\right)^{2/3} = 125^{2/3} = \left(\sqrt[3]{125}\right)^{2} = 5^{2} = 25$

Check it: $25^{3/2} = \left(\sqrt{25}\right)^{3} = 5^{3} = 125$.

A power of a power multiplies the exponents, and these two exponents are reciprocals.

$\left(a^{2/5}\right)^{5/2} = a^{\frac25 \cdot \frac52} = a^{1} = a$

Raising to $\tfrac{n}{m}$ undoes raising to $\tfrac{m}{n}$, which is exactly how you solve an equation like $x^{2/5} = c$ for a positive $x$.

Simplify what is under the cube root first. Multiplying powers of $x$ adds the exponents.

$x^{2} \cdot x^{1/2} = x^{5/2}$

Now apply the cube root, which is the power $\tfrac13$, so the exponents multiply.

$\left(x^{5/2}\right)^{1/3} = x^{\frac52 \cdot \frac13} = x^{5/6}$

Evaluate the two powers separately, letting each exponent act on the numerator and the denominator.

$\left(\frac{27}{8}\right)^{2/3} = \left(\frac{3}{2}\right)^{2} = \frac{9}{4}, \qquad \left(\frac{16}{9}\right)^{1/2} = \frac{4}{3}$

Now subtract over the common denominator $12$.

$\frac{27}{12} - \frac{16}{12} = \frac{11}{12}$

A fourth power is an even power, and an even power of any real number is never negative.

$2^{4} = 16 \quad \text{and} \quad (-2)^{4} = 16, \quad \text{so no real } r \text{ has } r^{4} = -16$

So the expression names no real number. Odd roots of negatives, such as $(-8)^{1/3} = -2$, are a different matter; only odd indices survive a negative radicand in the real numbers.

The argument shows that $\left(\sqrt[n]{a}\right)^{m}$ is a positive number whose $n$-th power is $a^{m}$, and then it needs to conclude that this pins the number down uniquely.

$\left(r^{m}\right)^{n} = r^{mn} = \left(r^{n}\right)^{m} = a^{m} \quad \text{with} \quad r = \sqrt[n]{a} > 0$

Only the uniqueness of the positive $n$-th root turns that into an equality of names. The claim that every real number has exactly one $n$-th root is false, since $4$ has two real square roots, and it is exactly that failure of uniqueness that the positivity requirement repairs.

Rewrite each factor as a power and simplify it on its own.

$\left(2^{1/2}\right)^{8} = 2^{4} = 16, \qquad 4^{-3/2} = \frac{1}{\left(\sqrt{4}\right)^{3}} = \frac{1}{8}$

Now multiply.

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

Each individual computation is arithmetically fine: $(-64)^{2} = 4096$ and $4096^{1/6} = 4$, while $(-4)^{3} = -64$ gives $(-64)^{1/3} = -4$.

$\left((-64)^{2}\right)^{1/6} = 4 \quad \text{but} \quad (-64)^{1/3} = -4$

An expression names one number, so the fault is in the notation, not the arithmetic. Well-definedness of $a^{m/n}$ was proved from the fact that a positive base has exactly one positive $n$-th root, and $-64$ has none. For a positive base both routes agree, and no such conflict can arise.

Combine the denominator first by adding its exponents.

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

The denominator matches the numerator exactly, so dividing subtracts equal exponents.

$\frac{x^{5/6}}{x^{5/6}} = x^{0} = 1$