Rational Exponents: Free Response

The denominator of each exponent names the index of the root, and the numerator names the power applied afterward.

For $625^{3/4}$ the index is $4$, and $5^{4} = 625$, so

$625^{3/4} = \left(\sqrt[4]{625}\right)^{3} = 5^{3} = 125.$

For $1000^{-2/3}$, deal with the minus sign first: it asks for a reciprocal, not a negative value. The remaining exponent $\tfrac23$ has index $3$, and $10^{3} = 1000$, so

$1000^{-2/3} = \frac{1}{1000^{2/3}} = \frac{1}{\left(\sqrt[3]{1000}\right)^{2}} = \frac{1}{10^{2}} = \frac{1}{100}.$

Translate one feature at a time: the index of the root becomes the denominator of the exponent, and the power inside becomes the numerator.

The fifth root of $p^{2}$ has index $5$ and inside power $2$, so

$\sqrt[5]{p^{2}} = p^{2/5}.$

Running the dictionary the other way, the minus sign in $p^{-4/9}$ asks for a reciprocal first, and the exponent $\tfrac49$ that remains is a ninth root carrying the power $4$:

$p^{-4/9} = \frac{1}{p^{4/9}} = \frac{1}{\sqrt[9]{p^{4}}}.$

Compare the two routes for $625^{3/4}$. Root first: $\sqrt[4]{625}=5$, then $5^{3}=125$, arithmetic that never leaves single digits until the last step. Power first, the base is cubed before any root is taken:

$625^{3} = 244{,}140{,}625,$

and only then would a fourth root of that nine-digit number need to be taken.

Both routes must agree, since $625^{3/4}$ names one number, but one of them is far easier to carry out by hand.

The rule read off the notation itself does not depend on which route is taken: in $a^{m/n}$, the denominator $n$ is the index of the root and the numerator $m$ is the power, and computing the root first is simply the version of that instruction that keeps the arithmetic small.

Multiplying same-base powers adds the exponents, and dividing subtracts them, exactly as for whole numbers.

For the product, put both exponents over the common denominator $6$:

$a^{5/6}\cdot a^{-1/3} = a^{\frac56 - \frac26} = a^{3/6} = a^{1/2}.$

For the quotient, dividing by $b^{-1/8}$ subtracts $-\tfrac18$, and subtracting a negative adds:

$\frac{b^{3/4}}{b^{-1/8}} = b^{\frac34 - \left(-\frac18\right)} = b^{\frac68 + \frac18} = b^{7/8}.$

Combine what is inside the parentheses first. Dividing by $m^{1/5}$ subtracts that exponent from $m$'s:

$\frac{m^{-2/5}}{m^{1/5}} = m^{-\frac25 - \frac15} = m^{-3/5}.$

The $n$ term has nothing to combine with, so the bracket is $m^{-3/5}n^{3/10}$. Raising to the $-10$ multiplies every exponent by $-10$:

$\left(m^{-3/5}n^{3/10}\right)^{-10} = m^{\left(-\frac35\right)(-10)}\,n^{\left(\frac{3}{10}\right)(-10)} = m^{6}n^{-3}.$

Clearing the negative exponent moves that factor to the denominator: $m^{6}n^{-3} = \dfrac{m^{6}}{n^{3}}$.

The lesson is explicit that $a^{1/n}$ is not a count of copies of $a$; it is the single number forced by the power rule and positivity. So a picture of 'groups of factors' cannot be the literal reason the fractional laws hold. What actually licenses them is the well-definedness argument from earlier in the lesson: any rational exponent can be rewritten as an integer power of one common root.

Every exponent in part B's bracket, $m^{-2/5}n^{3/10}/m^{1/5}$, has a denominator dividing $10$. Let $s=m^{1/10}$ and $u=n^{1/10}$, both positive real numbers since $m,n>0$. Multiplying each fractional exponent by $10$ recovers an integer exponent on $s$ or $u$:

$m^{-2/5}=s^{-4}, \qquad n^{3/10}=u^{3}, \qquad m^{1/5}=s^{2}.$

Now the bracket is $s^{-4}u^{3}/s^{2}=s^{-6}u^{3}$, and every exponent in sight is a whole number: combining $-4$ copies of $s$ and removing $2$ more really is subtraction, and raising to the outer $-10$ really is integer repetition, $(s^{-6}u^{3})^{-10}=s^{60}u^{-30}$. Translating back, $s^{60}=\left(m^{1/10}\right)^{60}=m^{6}$ and $u^{-30}=\left(n^{1/10}\right)^{-30}=n^{-3}$, matching part B.

So the addition-for-product, multiplication-for-power pattern is not a separate rule invented for fractions; it is the ordinary integer pattern, true by literal counting on $s$ and $u$, translated back through the division by $10$ that defined them.

Distributing the outer power first confirms the same answer without appealing to $s$ or $u$ at all: $\left(m^{-2/5}\right)^{-10} = m^{4}$, $\left(n^{3/10}\right)^{-10} = n^{-3}$, and $\left(m^{1/5}\right)^{-10} = m^{-2}$ in the denominator. Dividing by $m^{-2}$ multiplies by $m^{2}$, so the two powers of $m$ combine as $m^{4}\cdot m^{2} = m^{6}$, leaving $m^{6}n^{-3}$ once again.

The exponent $\tfrac32$ has index $2$, so take the square root of the distance first, then cube the result:

$T = 25^{3/2} = \left(\sqrt{25}\right)^{3} = 5^{3} = 125 \text{ years}.$

Raise both sides of $T=a^{3/2}$ to the power that undoes $\tfrac32$, which is its reciprocal $\tfrac23$:

$T^{2/3} = \left(a^{3/2}\right)^{2/3} = a^{1} = a.$

So $a=T^{2/3}$. For $T=216$, the index is $3$, and $6^{3}=216$:

$a = 216^{2/3} = \left(\sqrt[3]{216}\right)^{2} = 6^{2} = 36 \text{ AU}.$

Replace $a$ with $3a$ in the formula and pull the constant factor of $3$ out through the power, which is what $(ka)^{r}=k^{r}a^{r}$ licenses:

$T_X = (3a)^{3/2} = 3^{3/2}\,a^{3/2} = 3^{3/2}\,T_Y.$

So X's period is $3^{3/2}$ times Y's, and

$3^{3/2} = \left(\sqrt3\right)^{3} = 3\sqrt3 \approx 5.20.$

The factor is not $3$ because the same exponent $\tfrac32$ that acts on the whole distance acts on the scaling factor $3$ as well; a distance three times as large produces a period more than three times as large, since the exponent exceeds $1$.

Route 1 reduces the fraction before doing anything else. Since $\gcd(2,10)=2$, $\tfrac{2}{10}=\tfrac15$, and a fifth root of a negative number is a perfectly good real number, negative, since $5$ is odd:

$(-243)^{1/5} = -3, \quad \text{because } (-3)^{5} = -243.$

Route 2 leaves the fraction as it is and squares the base first, an even power that erases the sign:

$(-243)^{2} = 59049 = 3^{10},$

and a tenth root of a positive number is the positive number that raises to it:

$59049^{1/10} = 3.$

The two routes, both built from rules used elsewhere in this lesson, hand back different numbers: $-3$ and $3$.

A tenth root of $-243$ would be a real number $x$ solving

$x^{10} = -243.$

An even power of any real number, positive or negative, is never negative, so no such $x$ exists. This route hands back nothing at all, not even a candidate.

So the three routes give three different outcomes: $-3$, $3$, and no real number. Every manipulation used to reach the first two, rewriting a fraction, moving a power inside a root, was proved in this lesson only for a positive base. With $a=-243$ that hypothesis fails, and once it fails there is no longer any guarantee that different legal-looking manipulations land on the same answer; here they demonstrably do not.

With $\tfrac{m}{n}$ already in lowest terms, there is no fraction left to reduce and no ambiguity about which route to take:

$a^{m/n} = \left(a^{1/n}\right)^{m}, \quad \text{root then power, with nothing else to choose between.}$

If $n$ is odd, $a^{1/n}$ is a genuine real number for any real $a$, negative included, and raising a real number to the integer power $m$ stays real, so the whole expression is real.

If $n$ is even, $a^{1/n}$ is not real when $a<0$, since no real number raised to an even power is negative. There is then nothing for the outer power $m$ to act on, so the expression fails to exist before $m$ is even considered.

The rule: for $a<0$ and $\tfrac{m}{n}$ in lowest terms, $a^{m/n}$ is real exactly when $n$ is odd.

For $(-16)^{5/2}$, the fraction $\tfrac52$ is already in lowest terms and $n=2$ is even, so $(-16)^{5/2}$ is not a real number, whatever value $(-16)^{5}$ itself takes.

Let $x=a^{1/4}$, whatever that value turns out to be. Assumption (i) says

$x^{4} = \left(a^{1/4}\right)^{4} = a^{\frac14\cdot4} = a^{1} = a,$

so $x$ is some fourth root of $a$. Assumption (ii) says $x>0$, so $x$ is specifically a positive fourth root of $a$.

The result already established in this lesson is that a positive number has exactly one positive $n$th root, for any index $n$, $4$ included. So there is exactly one positive number whose fourth power is $a$, and its name is $\sqrt[4]{a}$. Since $x$ is a positive fourth root of $a$ and only one exists, $x=\sqrt[4]{a}$, and no other real number could have satisfied both assumptions at once.

Since $3^{4}=81$, the fourth root of $81$ is $3$, so

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

For $81^{3/4}$, take that same fourth root first, then cube it:

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

The equation

$x^{4} = a$

is what assumption (i) forces $x$ to satisfy. When $a<0$, that equation has no real solution whatsoever, since a fourth power of any real number, positive or negative, is never negative. So it is not that assumption (ii) rules out a negative candidate for $x$; there is no real candidate at all, positive or negative, for assumption (ii) to apply to.

The breakdown happens one step earlier than it might seem: assumption (i) alone already has nothing real to hand over once $a<0$, so nothing is left for the uniqueness lemma to identify. That is exactly why this lesson restricts $a^{1/n}$ to $a>0$ before any of its results are stated.