Rational exponents
An exponent counts how many times a number is used as a factor. What could it mean to use 1331 as a factor one third of a time?
Where you are 5 in Exponents and roots
Why this exists
A whole-number exponent counts factors: 11^3 = 11 \cdot 11 \cdot 11 = 1331. A count of factors can be three or four, but it cannot be one third. Yet fractional exponents turn up as soon as questions are asked in reverse, or as soon as a formula from science is written down. The time a planet takes to go round the Sun, for instance, measured in suitable units, is its average distance from the Sun raised to the power \tfrac32 (Example 8).
The way forward is the one already taken for the zero exponent: instead of asking what the symbol counts, ask what it has to equal if the laws of exponents are to keep working. The power of a power law says (a^m)^n = a^{mn}. If that law is to hold when m = \tfrac13, then
So 1331^{1/3} has to be a number whose cube is 1331. There is exactly one real number like that, the cube root \displaystyle \sqrt[3]{1331} = 11. Nobody chose this meaning. The law forced it, and with it every fraction gets a meaning as an exponent.
Definition
Unit fractions. For a whole number n \ge 2, the power a^{1/n} is the nth root of a:
It is defined exactly when \displaystyle \sqrt[n]{a} is a real number: for every real a when n is odd, and for a \ge 0 when n is even. So \displaystyle a^{1/2} = \sqrt{a} and \displaystyle a^{1/3} = \sqrt[3]{a}: the denominator of the exponent is the index of the root.
Any fraction. Let \tfrac mn be a fraction in lowest terms, with m an integer and n \ge 2 a whole number. Then
defined whenever \displaystyle \sqrt[n]{a} is real, except that a = 0 is not allowed when m is negative. The denominator says which root to take and the numerator says which power of that root. An exponent written as a fraction of two integers is a rational exponent. When the exponent is an integer (a fraction with denominator 1), it means what it always meant.
Because m may be negative, the old meaning of a negative exponent carries over unchanged: for a \ne 0,
The words “lowest terms” matter only when the base is negative, and the Boundaries section shows why they are needed there. For a positive base, every way of writing the fraction gives the same value (Results).
As always, an exponent belongs only to what it is written on. In -1331^{1/3} the exponent is on 1331, and the minus sign is applied afterwards: -1331^{1/3} = -11.
Intuition
Think of 8^1 as one step that multiplies by 8, starting from 8^0 = 1. Now cut that step into three equal steps. Each small step must multiply by the same factor r, and three of them together must multiply by 8, so r \cdot r \cdot r = 8 and r = 2. The heights along the way are 8^{1/3} = 2 and 8^{2/3} = 2 \cdot 2 = 4, and the third step lands on 8.
That is the whole idea of a^{m/n}. The denominator n cuts one full factor of a into n equal smaller factors, each equal to \displaystyle \sqrt[n]{a}, and the numerator m says how many of those smaller factors to take. Taking m of them can be done in either order: take the root first and then the power, or the power first and then the root. Both give the same number (Results), but taking the root first keeps the numbers small, which is why it is usually the better way to work by hand.
The figure also shows something larger. Fractional exponents fill in the gaps between the whole numbers, so the dots join up into an unbroken curve, y = 8^x. That curve is an exponential function, and it could not even be drawn without this definition.
Results
Unit fraction exponents. For a whole number n \ge 2, whenever \displaystyle \sqrt[n]{a} is real,
Rational exponents. For \tfrac mn in lowest terms with n \ge 2, whenever \displaystyle \sqrt[n]{a} is real (and a \ne 0 if m < 0), the root and the power can be taken in either order:
Negative rational exponents. For a \ne 0, whenever a^{m/n} is real, a^{-m/n} = \dfrac{1}{a^{m/n}}.
Equal fractions give equal powers. For a > 0 and any whole number k \ge 1, a^{km/(kn)} = a^{m/n}. So for a positive base the fraction need not be in lowest terms.
The laws of exponents still hold. For a > 0, b > 0 and any rational exponents r and s:
Why it is true
Why a^{1/n} is the root and not anything else. If the power of a power law is to hold, then \left(a^{1/n}\right)^n = a^1 = a, so a^{1/n} has to be a number whose nth power is a. When n is odd there is exactly one real number like that. When n is even and a > 0 there are two, \displaystyle \sqrt[n]{a} and \displaystyle -\sqrt[n]{a}, and the law itself picks the positive one: by the same law, a^{1/n} = \left(a^{1/(2n)}\right)^2, a square, which cannot be negative.
Why the two orders agree. Let \displaystyle r = \sqrt[n]{a}, a real number. Using only the whole-number laws, together with the commutative property of multiplication (mn = nm),
So r^m is a number whose nth power is a^m. When n is odd that number is unique, so \displaystyle r^m = \sqrt[n]{a^m}. When n is even, a \ge 0, so r \ge 0 and r^m \ge 0; it is the non-negative nth root of a^m, which is again \displaystyle \sqrt[n]{a^m}.
Why equal fractions give equal powers. Let a > 0 and let \displaystyle s = \sqrt[kn]{a}, which is positive. Then \left(s^k\right)^n = s^{kn} = a and s^k > 0, so s^k is the positive nth root of a: \displaystyle s^k = \sqrt[n]{a}. Therefore
Why the laws still hold. Take a > 0. Any two rational exponents can be written over a common denominator, r = \tfrac pn and s = \tfrac qn, and by the last result this does not change either power. Let \displaystyle t = \sqrt[n]{a}. Then, using the whole-number product law on the base t,
The quotient law follows the same way, with t^{p-q}. For the power of a power law, the same reasoning first shows that a root of a power is a power with the fraction divided: \left(a^{p/n}\right)^{1/k} = a^{p/(nk)}, because a^{p/(nk)} is positive and, by the product law used k times, its kth power is a^{kp/(nk)} = a^{p/n}. Raising that to a whole-number power q multiplies the exponent by q, which gives \left(a^{p/n}\right)^{q/k} = a^{pq/(nk)}. The two laws for products and quotients of bases follow from \displaystyle \sqrt[n]{ab} = \sqrt[n]{a}\,\sqrt[n]{b} in the same way.
Every step above leans on the base being positive. Boundaries shows a negative base breaking the laws.
Worked examples
Powers of one third and one fourth
Evaluate each, or say why it is not a real number: 1331^{1/3}, 14641^{1/4}, (-1331)^{1/3}, (-14641)^{1/4} and -14641^{1/4}.
SolutionEach power of \tfrac1n is an nth root, so each question asks for a number that, raised to the power n, gives the base.
\displaystyle 1331^{1/3} = \sqrt[3]{1331} = 11, since 11^3 = 1331.
\displaystyle 14641^{1/4} = \sqrt[4]{14641} = 11, since 11^4 = 121^2 = 14641. An even root has two candidates, 11 and -11, and the root is the positive one.
\displaystyle (-1331)^{1/3} = \sqrt[3]{-1331} = -11, since (-11)^3 = -1331. An odd root accepts a negative base, because an odd power of a negative number is negative.
(-14641)^{1/4} is not a real number. A fourth power is the square of a square, so it is never negative, and no real number has fourth power -14641.
-14641^{1/4} = -\left(14641^{1/4}\right) = -11. The exponent is on 14641 alone, and the minus sign comes afterwards, so this one is real.
Between radicals and exponents
Write z^{1/5}, (6y)^{1/3} and 6y^{1/3} as radicals, and write \displaystyle \sqrt[4]{5z}, \displaystyle \sqrt[5]{z^2} and \displaystyle \dfrac{1}{\sqrt{z}} with rational exponents.
SolutionThe denominator of the exponent is the index of the root, and the numerator is the power.
\displaystyle z^{1/5} = \sqrt[5]{z}.
\displaystyle (6y)^{1/3} = \sqrt[3]{6y}, because the brackets put the whole product under the exponent. But \displaystyle 6y^{1/3} = 6\sqrt[3]{y}, because without brackets the exponent belongs to y alone.
\displaystyle \sqrt[4]{5z} = (5z)^{1/4}. The brackets are needed, for the same reason.
\displaystyle \sqrt[5]{z^2} = z^{2/5}: a fifth root of the square, so denominator 5 and numerator 2.
\displaystyle \dfrac{1}{\sqrt{z}} = \dfrac{1}{z^{1/2}} = z^{-1/2}, since one over a power is the power with the opposite exponent.
Root first or power first
Evaluate 1331^{2/3} both ways, then evaluate 16^{5/4}, \left(\tfrac49\right)^{3/2} and (-27)^{2/3}.
SolutionRoot first: \displaystyle 1331^{2/3} = \left(\sqrt[3]{1331}\right)^2 = 11^2 = 121.
Power first: \displaystyle 1331^{2/3} = \sqrt[3]{1331^2} = \sqrt[3]{1771561}. The cube root of a seven-digit number is hard to see, but 121^3 = 1771561, so it is 121 again. The two orders agree, as the Results promised, and the root-first order was far easier.
\displaystyle 16^{5/4} = \left(\sqrt[4]{16}\right)^5 = 2^5 = 32.
\displaystyle \left(\tfrac49\right)^{3/2} = \left(\sqrt{\tfrac49}\right)^3 = \left(\tfrac23\right)^3 = \tfrac{8}{27}. The root of a fraction is the root of the top over the root of the bottom.
\displaystyle (-27)^{2/3} = \left(\sqrt[3]{-27}\right)^2 = (-3)^2 = 9. The base is negative, but the denominator 3 is odd, so the cube root is real, and squaring it gives a positive answer.
Negative rational exponents
Evaluate 1331^{-2/3}, \left(\tfrac{8}{27}\right)^{-2/3}, -16^{-3/4} and (-32)^{-3/5}.
SolutionA negative exponent means one over the positive power, and the positive power is found as in the last example.
1331^{-2/3} = \dfrac{1}{1331^{2/3}} = \dfrac{1}{121}.
For a fraction, one over the power is the power of the flipped fraction: \left(\tfrac{8}{27}\right)^{-2/3} = \left(\tfrac{27}{8}\right)^{2/3} = \left(\tfrac32\right)^2 = \tfrac94.
-16^{-3/4} = -\dfrac{1}{16^{3/4}} = -\dfrac{1}{2^3} = -\dfrac18. The minus sign is outside the power and stays outside; it does not make the exponent change.
(-32)^{-3/5} = \dfrac{1}{(-32)^{3/5}} = \dfrac{1}{(-2)^3} = -\dfrac18, since (-2)^5 = -32. The denominator 5 is odd, so the negative base is allowed.
Using the laws
Simplify, taking every variable to be positive: (a) x^{1/2} \cdot x^{1/3}; (b) \left(y^{3/4}\right)^{2/3}; (c) \left(27m^6 n^{-3/2}\right)^{2/3}; (d) \dfrac{p^{5/4}}{p^{-1/4}}.
SolutionThe laws hold for positive bases, so they can be used exactly as with whole-number exponents. The only new work is adding and multiplying fractions.
(a) Same base, multiplied, so the exponents add: \tfrac12 + \tfrac13 = \tfrac36 + \tfrac26 = \tfrac56, and x^{1/2} \cdot x^{1/3} = x^{5/6}.
(b) A power of a power, so the exponents multiply: \tfrac34 \cdot \tfrac23 = \tfrac{6}{12} = \tfrac12, and \left(y^{3/4}\right)^{2/3} = y^{1/2}.
(c) A power of a product is the product of the powers, so the exponent \tfrac23 goes onto each factor:
Here 27^{2/3} = 3^2 = 9.
(d) Same base, divided, so the exponents subtract: \tfrac54 - \left(-\tfrac14\right) = \tfrac64 = \tfrac32, and the quotient is p^{3/2}.
Joining roots of different index
Taking x > 0, write each as a single radical: (a) \displaystyle \sqrt[6]{x^4}; (b) \displaystyle \sqrt{x} \cdot \sqrt[3]{x}; (c) \displaystyle \sqrt[3]{\sqrt{x}}. Then (d) write \displaystyle \sqrt2 \cdot \sqrt[3]{2} as a single root of a whole number.
SolutionTwo roots with different indexes cannot be joined while they are written as roots. Written as powers, they obey the laws, and the answer can then be turned back into a root.
(a) \displaystyle \sqrt[6]{x^4} = x^{4/6} = x^{2/3} = \sqrt[3]{x^2}. Reducing the fraction lowered the index of the root from 6 to 3. This is allowed because x > 0.
(b) \displaystyle \sqrt{x} \cdot \sqrt[3]{x} = x^{1/2} \cdot x^{1/3} = x^{5/6} = \sqrt[6]{x^5}, using the sum found in Example 5.
(c) \displaystyle \sqrt[3]{\sqrt{x}} = \left(x^{1/2}\right)^{1/3} = x^{1/6} = \sqrt[6]{x}. A cube root of a square root is a sixth root.
(d) \displaystyle \sqrt2 \cdot \sqrt[3]{2} = 2^{1/2} \cdot 2^{1/3} = 2^{5/6} = \sqrt[6]{2^5} = \sqrt[6]{32}.
Check (d). Raise both sides to the sixth power. On the left, \left(2^{1/2}\right)^6 \cdot \left(2^{1/3}\right)^6 = 2^3 \cdot 2^2 = 32; on the right, 32.
Solving with rational exponents
Solve (a) x^{3/2} = 1331 and (b) x^{2/3} = 121.
SolutionWrite each power as a root and a power, then undo them one at a time.
(a) \displaystyle x^{3/2} = \left(\sqrt{x}\right)^3, which needs x \ge 0. So \displaystyle \left(\sqrt{x}\right)^3 = 1331. Only one real number has cube 1331, so \displaystyle \sqrt{x} = 11, and squaring gives x = 121.
Check. \displaystyle 121^{3/2} = \left(\sqrt{121}\right)^3 = 11^3 = 1331.
(b) \displaystyle x^{2/3} = \left(\sqrt[3]{x}\right)^2, and the denominator 3 is odd, so here x may be negative. The equation says \displaystyle \left(\sqrt[3]{x}\right)^2 = 121, so \displaystyle \sqrt[3]{x} is a number whose square is 121: either 11 or -11. Cubing, x = 1331 or x = -1331.
Check. 1331^{2/3} = 11^2 = 121 and (-1331)^{2/3} = (-11)^2 = 121.
A shortcut for (b) would be to raise both sides to the power \tfrac32, giving x = 121^{3/2} = 1331. It finds only one of the two solutions, because the law \left(x^{2/3}\right)^{3/2} = x holds only for x \ge 0.
Kepler’s third law
For a body going round the Sun, measure its period T (the time for one orbit) in years and its average distance a from the Sun in astronomical units, where one astronomical unit is the Earth’s average distance from the Sun. Then, to good accuracy, T^2 = a^3. (a) Jupiter’s average distance is about 5.2 astronomical units. Find its period. (b) A comet has a period of 1331 years. Find its average distance.
SolutionBoth quantities are positive, so the laws of exponents may be used freely. Raising both sides of T^2 = a^3 to the power \tfrac12 gives T = a^{3/2}, and raising them to the power \tfrac13 gives a = T^{2/3}.
(a) \displaystyle T = 5.2^{3/2} = 5.2 \cdot \sqrt{5.2} \approx 5.2 \cdot 2.280 \approx 11.86. Jupiter takes about 11.9 years to go round the Sun. Here 5.2^{3/2} = 5.2^{1} \cdot 5.2^{1/2} by the product law.
(b) \displaystyle a = 1331^{2/3} = \left(\sqrt[3]{1331}\right)^2 = 11^2 = 121. The comet’s average distance is 121 astronomical units.
Check (b). T^2 = 1331^2 = 1771561 and a^3 = 121^3 = 1771561.
Boundaries
An even root of a negative number is not real. a^{1/n} with n even is defined only for a \ge 0, so (-16)^{1/4} and (-14641)^{1/4} are not real numbers. The same holds for a^{m/n} whenever the denominator n, in lowest terms, is even.
A negative base needs the fraction in lowest terms. Without that rule, (-8)^{1/3} and (-8)^{2/6}, the same exponent written two ways, would give different answers: (-8)^{1/3} = -2, but working from \tfrac26 as written gives \displaystyle \sqrt[6]{(-8)^2} = \sqrt[6]{64} = 2. The definition reduces \tfrac26 to \tfrac13 first, so (-8)^{2/6} = -2.
The laws of exponents need a positive base. With a negative base they can fail even when every power is real: \left((-1)^2\right)^{1/2} = 1^{1/2} = 1, while \displaystyle (-1)^{2 \cdot \frac12} = (-1)^1 = -1. This is why simplifying with the laws is done for positive variables, and why the shortcut in Example 7 lost a solution.
Zero. 0^{m/n} = 0 when \tfrac mn > 0, and 0^{m/n} is undefined when \tfrac mn < 0, since it would mean dividing by 0.
Not every book agrees about negative bases. Some texts, and many calculators and computer languages, define a^{m/n} only for a > 0, or return a complex number for a power such as (-8)^{1/3}. The definition here, with the real odd root, is the one used in this lesson, and it is worth checking which one a calculator uses before trusting it with a negative base.
Rational exponents only. The definition covers exponents that are fractions of integers. A power such as \displaystyle 2^{\sqrt2}, with an irrational exponent, is not covered; it is defined later by approximating \displaystyle \sqrt2 by fractions, which is part of the study of exponential functions.
Where this goes next
Rational exponents turn every root into a power, so the work of radicals and rational expressions can be done with the laws of exponents: simplifying radicals and combining roots of different index, as in Example 6, become arithmetic with fractions. Equations with roots, studied in solving radical equations, are the same kind of equation as Example 7. The graphs of the roots as powers, y = x^{1/n}, are met in cube root functions.
Further on, the exponent stops being a fixed number and becomes the variable, which turns 8^x into an exponential function. And in calculus, the power rule for x^n extends to every rational exponent, so \displaystyle \sqrt{x} = x^{1/2} is handled by the same rule as x^2.
Exercises
Unit fraction exponents
- Evaluate 169^{1/2}, 1000^{1/3} and 0.0001^{1/4}.
- Evaluate each, or say why it is not a real number: (-512)^{1/3}, (-64)^{1/2}, -64^{1/2} and (-64)^{1/3}.
- Evaluate \left(\tfrac{1}{81}\right)^{1/4} and \left(-\tfrac{1}{32}\right)^{1/5}.
Between radicals and exponents
- Write as a radical: m^{1/4}, (3m)^{1/2}, 3m^{1/2} and m^{3/5}.
- Write with a rational exponent: \displaystyle \sqrt[3]{k}, \displaystyle \sqrt[5]{k^4}, \displaystyle \dfrac{1}{\sqrt[3]{k^2}} and \displaystyle \left(\sqrt[4]{2k}\right)^3.
Fractional powers
- Evaluate 64^{2/3}, 9^{5/2} and 0.01^{3/2}.
- Evaluate 81^{3/4}, (-125)^{2/3} and (-8)^{5/3}.
- Evaluate \left(\tfrac{16}{81}\right)^{3/4}.
- Evaluate 4^{3/2} twice, once taking the root first and once taking the power first.
Negative rational exponents
- Evaluate 49^{-1/2} and 8^{-2/3}.
- Evaluate \left(\tfrac{1}{27}\right)^{-4/3} and -36^{-3/2}.
- Evaluate \left(\tfrac{25}{4}\right)^{-3/2}.
Using the laws
In Exercises 13 to 18, every variable is positive.
- Simplify x^{2/3} \cdot x^{1/6}.
- Simplify \left(z^{4/5}\right)^{5/8}.
- Simplify \dfrac{w^{3/4}}{w^{1/3}}.
- Simplify \left(16a^8 b^{-2}\right)^{3/4}, and write the answer with no negative exponent.
- Simplify \left(x^{1/2} y^{1/3}\right)^6.
- Write each as a single radical: \displaystyle \sqrt{x} \cdot \sqrt[4]{x} and \displaystyle \sqrt[3]{\sqrt[4]{x}}.
Equations and uses
- Solve x^{1/3} = 5 and x^{3/4} = 8.
- Solve x^{2/3} = 49.
- Using T^2 = a^3 from Example 8, find the period of a body whose average distance from the Sun is 16 astronomical units, and the average distance of a body whose period is 343 years.
Reasoning
- Ines says that 27^{1/3} ought to mean 27 \div 3 = 9. Use the power of a power law to explain why 27^{1/3} has to be 3 instead.
- Explain why (-128)^{1/7} is a real number but (-81)^{1/4} is not, and find the one that is real.
- Owen writes \displaystyle (-8)^{2/6} = \sqrt[6]{(-8)^2} = \sqrt[6]{64} = 2. Explain what the definition gives instead, and why the definition is written that way.
- Show from the definition that a^{2/4} = a^{1/2} for every a > 0.
Answers
- 13, 10 and 0.1. Since 13^2 = 169, 10^3 = 1000 and 0.1^4 = 0.0001.
- (-512)^{1/3} = -8, since (-8)^3 = -512. (-64)^{1/2} is not a real number, since no real square is negative. \displaystyle -64^{1/2} = -\sqrt{64} = -8, since the exponent is on 64 alone. (-64)^{1/3} = -4, since (-4)^3 = -64.
- \tfrac13 and -\tfrac12. Since 3^4 = 81, \displaystyle \sqrt[4]{\tfrac{1}{81}} = \tfrac13; since (-2)^5 = -32, \displaystyle \sqrt[5]{-\tfrac{1}{32}} = -\tfrac12.
- \displaystyle \sqrt[4]{m}, \displaystyle \sqrt{3m}, \displaystyle 3\sqrt{m} and \displaystyle \sqrt[5]{m^3} (or \displaystyle \left(\sqrt[5]{m}\right)^3). In 3m^{1/2} the exponent is on m alone.
- k^{1/3}, k^{4/5}, k^{-2/3} and (2k)^{3/4}. The index becomes the denominator, the power the numerator, and one over a power is a negative exponent.
- 16, 243 and 0.001. \displaystyle \left(\sqrt[3]{64}\right)^2 = 4^2 = 16; \displaystyle \left(\sqrt9\right)^5 = 3^5 = 243; \displaystyle \left(\sqrt{0.01}\right)^3 = 0.1^3 = 0.001.
- 27, 25 and -32. \displaystyle \left(\sqrt[4]{81}\right)^3 = 3^3 = 27; \displaystyle \left(\sqrt[3]{-125}\right)^2 = (-5)^2 = 25; \displaystyle \left(\sqrt[3]{-8}\right)^5 = (-2)^5 = -32.
- \tfrac{8}{27}. \displaystyle \sqrt[4]{\tfrac{16}{81}} = \tfrac23, and \left(\tfrac23\right)^3 = \tfrac{8}{27}.
- 8 both ways. Root first: \displaystyle \left(\sqrt4\right)^3 = 2^3 = 8. Power first: \displaystyle \sqrt{4^3} = \sqrt{64} = 8.
- \tfrac17 and \tfrac14. \displaystyle 49^{-1/2} = \dfrac{1}{\sqrt{49}} = \tfrac17; \displaystyle 8^{-2/3} = \dfrac{1}{\left(\sqrt[3]{8}\right)^2} = \dfrac{1}{2^2} = \tfrac14.
- 81 and -\tfrac{1}{216}. \left(\tfrac{1}{27}\right)^{-4/3} = 27^{4/3} = 3^4 = 81; \displaystyle -36^{-3/2} = -\dfrac{1}{\left(\sqrt{36}\right)^3} = -\dfrac{1}{6^3} = -\tfrac{1}{216}.
- \tfrac{8}{125}. Flip the fraction: \left(\tfrac{4}{25}\right)^{3/2} = \left(\tfrac25\right)^3 = \tfrac{8}{125}.
- x^{5/6}. The exponents add: \tfrac23 + \tfrac16 = \tfrac46 + \tfrac16 = \tfrac56.
- z^{1/2}. The exponents multiply: \tfrac45 \cdot \tfrac58 = \tfrac{20}{40} = \tfrac12.
- w^{5/12}. The exponents subtract: \tfrac34 - \tfrac13 = \tfrac{9}{12} - \tfrac{4}{12} = \tfrac{5}{12}.
- \dfrac{8a^6}{b^{3/2}}. The power goes onto each factor: 16^{3/4} = 2^3 = 8, \displaystyle a^{8 \cdot \frac34} = a^6 and \displaystyle b^{-2 \cdot \frac34} = b^{-3/2} = \dfrac{1}{b^{3/2}}.
- x^3 y^2. \left(x^{1/2}\right)^6 = x^3 and \left(y^{1/3}\right)^6 = y^2.
- \displaystyle \sqrt[4]{x^3} and \displaystyle \sqrt[12]{x}. x^{1/2} \cdot x^{1/4} = x^{3/4}, and \left(x^{1/4}\right)^{1/3} = x^{1/12}.
- x = 125 and x = 16. \displaystyle \sqrt[3]{x} = 5 gives x = 5^3 = 125. \displaystyle \left(\sqrt[4]{x}\right)^3 = 8 needs x \ge 0; then \displaystyle \sqrt[4]{x} = 2, so x = 2^4 = 16.
- x = 343 or x = -343. \displaystyle \left(\sqrt[3]{x}\right)^2 = 49, so \displaystyle \sqrt[3]{x} = 7 or -7, and cubing gives 343 or -343. Both check: (\pm 7)^2 = 49.
- 64 years and 49 astronomical units. \displaystyle T = 16^{3/2} = \left(\sqrt{16}\right)^3 = 4^3 = 64; \displaystyle a = 343^{2/3} = \left(\sqrt[3]{343}\right)^2 = 7^2 = 49.
- If the power of a power law is to hold, \left(27^{1/3}\right)^3 = 27^1 = 27, so 27^{1/3} must be a number whose cube is 27. That number is 3. The value 9 fails the test, since 9^3 = 729, not 27.
- (-128)^{1/7} = -2, since (-2)^7 = -128: an odd power of a negative number is negative, so a negative number has a real odd root. (-81)^{1/4} would need a real number whose fourth power is -81, but a fourth power is the square of a square and is never negative.
- The definition reduces \tfrac26 to \tfrac13 first, so (-8)^{2/6} = (-8)^{1/3} = -2. Without that rule the same exponent, \tfrac26 = \tfrac13, would give two answers, 2 and -2, so the symbol would not have one value. For a positive base the rule makes no difference, which is why it is needed only for negative bases.
- Let \displaystyle s = \sqrt[4]{a} > 0. Then a^{2/4} = s^2. Also \left(s^2\right)^2 = s^4 = a and s^2 > 0, so s^2 is the positive square root of a. Hence \displaystyle a^{2/4} = s^2 = \sqrt{a} = a^{1/2}.