Laws of Exponents

(a) $32$   (b) $81$   (c) $-27$   (d) $1$   (e) ${\displaystyle \frac{8}{125}}$

Use the definition given at the start of this section: $b^n$ means $b$ multiplied by itself $n$ times.

(a) $2^5$. Multiply five $2$'s together, one step at a time:

So $2^5=32$. Note that $2^5$ is not $2\times 5=10$. The exponent counts how many factors there are, it is not itself a factor.

(b) $(-3{)}^4$. The parentheses tell us the base is $-3$, so we multiply four copies of $-3$:

So $(-3{)}^4=81$, a positive number. The minus signs cancel in pairs, and with four factors there are two such pairs.

(c) $(-3{)}^3$. Now there are only three copies:

So $(-3{)}^3=-27$, a negative number. One minus sign is left over without a partner.

The pattern behind (b) and (c) is worth remembering: a negative base raised to an even power gives a positive result, and to an odd power gives a negative result. This section uses that fact when it discusses odd and even indices.

(d) $7^0$. By the Zero Exponent Rule, any non-zero base raised to the power $0$ is $1$. Since $7\ne 0$,

(e) ${\left({\displaystyle \frac{2}{5}}\right)}^3$. By the Power of a Quotient Rule, ${\left({\displaystyle \frac{b}{c}}\right)}^n={\displaystyle \frac{b^n}{c^n}}$, so the exponent applies to the top and the bottom separately:

A common slip is to cube only the numerator and write ${\displaystyle \frac{8}{5}}$. The whole fraction is the base, so the whole fraction gets cubed.

(a) $x^{11}$   (b) $b^6$   (c) $y^{15}$   (d) $16t^4$   (e) ${\displaystyle \frac{a^2}{9}}$

Each part uses exactly one of the five laws. The skill is identifying which.

(a) $x^4\cdot x^7$. Two powers of the same base are being multiplied, so this is the Product Rule, $b^n\cdot b^m=b^{n+m}$. The exponents are added:

Why adding is right: $x^4$ is four $x$'s and $x^7$ is seven $x$'s, so together there are eleven $x$'s multiplied.

(b) ${\displaystyle \frac{b^9}{b^3}}$. Powers of the same base are being divided, so this is the Quotient Rule, ${\displaystyle \frac{b^n}{b^m}}=b^{n-m}$ (requiring $b\ne 0$). The exponents are subtracted:

(c) ${\left(y^3\right)}^5$. A power is being raised to another power, so this is the Power of a Power Rule, ${\left(b^n\right)}^m=b^{mn}$. The exponents are multiplied:

Parts (a) and (c) are the pair most often confused. Multiplying two powers adds exponents; raising a power to a power multiplies them. The bracket is the clue: no bracket means a product, a bracket means a power of a power.

(d) $(2t{)}^4$. The base is the product $2t$, so this is the Power of a Product Rule, $(bc{)}^n=b^n{c}^n$. The exponent goes onto each factor:

Do not leave the $2$ alone. Writing $2t^4$ would be wrong, since that is a different expression — test it with $t=1$, where $(2\cdot 1{)}^4=16$ but $2\cdot 1^4=2$.

(e) ${\left({\displaystyle \frac{a}{3}}\right)}^2$. The base is a quotient, so this is the Power of a Quotient Rule:

(a) $3$   (b) $2$   (c) $-4$   (d) $7$   (e) not a real number

By the definition in this section, $\sqrt[n]{b}=u$ means $u^n=b$. So in each part we hunt for a number that gives the value under the root when raised to the appropriate power, and then we check the sign rules.

(a) $\sqrt[3]{27}$. We need $u$ with $u^3=27$. Trying small numbers: $2^3=8$, too small; $3^3=27$, exactly right. So

(b) $\sqrt[4]{16}$. We need $u$ with $u^4=16$. Since $2^4=16$, the answer is $u=2$. Note that $(-2{)}^4=16$ as well, so there are two real fourth roots here. The index $4$ is even, and this section states that in that case the symbol $\sqrt[n]{}$ always denotes the positive (principal) root. So

not $-2$ and not "$\pm 2$".

(c) $\sqrt[3]{-64}$. We need $u$ with $u^3=-64$. An odd power keeps the sign of its base, so $u$ must be negative. Trying $u=-4$:

So $\sqrt[3]{-64}=-4$. This is allowed because the index $3$ is odd, and the section says every real number has exactly one real $n$th root when $n$ is odd.

(d) $\sqrt{49}$. The missing index is understood to be $2$. We need $u^2=49$, and $7^2=49$. The index is even, so we take the positive root:

(e) $\sqrt[4]{-16}$. We need $u$ with $u^4=-16$. But raising any real number to an even power gives a result that is zero or positive — a negative base has its minus signs cancel in pairs. So no real number can have fourth power $-16$, and

Compare (c) and (e) carefully. A negative number under an odd root is perfectly fine; under an even root it leaves the real numbers altogether. The index, not the minus sign alone, decides.

(a) $4$   (b) $5$   (c) $8$   (d) $81$

The Rational Exponent Rule says $b^{m/n}$ may be computed either as ${\left(b^{1/n}\right)}^m$ or as ${\left(b^m\right)}^{1/n}$. The two give the same answer, but taking the root first almost always keeps the numbers small, so that is the order used below.

A helpful way to read $m/n$: the bottom number tells you which root to take, and the top number tells you which power to raise it to.

(a) $8^{2/3}$. The denominator is $3$, so take the cube root first. Since $2^3=8$,

The numerator is $2$, so now square:

Check by the other order: $8^2=64$, and $\sqrt[3]{64}=4$ since $4^3=64$. Same answer, but it required cubing $8$ first.

(b) ${25}^{1/2}$. Here $m=1$, so this is just the square root:

(c) ${16}^{3/4}$. The denominator is $4$, so take the fourth root first. Since $2^4=16$,

The numerator is $3$, so cube it:

Check by the other order: ${16}^3=4096$, and $\sqrt[4]{4096}=8$ since $8^4=4096$. Same answer, with far more work.

(d) ${27}^{4/3}$. Cube root first: ${27}^{1/3}=3$, since $3^3=27$. Then raise to the fourth power:

The mistake to avoid throughout: do not confuse the roles of top and bottom. In $8^{2/3}$ the $3$ is the root and the $2$ is the power. Reading it the other way round would give ${\left(8^3\right)}^{1/2}=\sqrt{512}$, which is not even a whole number.

(a) ${\displaystyle \frac{1}{8}}$   (b) ${\displaystyle \frac{1}{5}}$   (c) ${\displaystyle \frac{9}{4}}$   (d) ${\displaystyle \frac{1}{2}}$

The Negative Exponent Rule says $b^{-r}={\displaystyle \frac{1}{b^r}}$ for $b\ne 0$. So a minus sign in the exponent means "take the reciprocal", and it says nothing at all about the sign of the answer.

(a) $2^{-3}$. Apply the rule with $b=2$ and $r=3$:

Note the answer is positive. Writing $-8$ would be wrong.

(b) $5^{-1}$.

An exponent of $-1$ always produces the reciprocal of the base.

(c) ${\left({\displaystyle \frac{2}{3}}\right)}^{-2}$. First apply the rule, taking the reciprocal of the whole thing:

Work out the bottom using the Power of a Quotient Rule:

So we need ${\displaystyle \frac{1}{4/9}}$. Dividing by a fraction means multiplying by its reciprocal:

So the answer is ${\displaystyle \frac{9}{4}}$. The useful shortcut visible here: a negative exponent on a fraction flips the fraction and then applies the positive exponent, since ${\left(\frac{2}{3}\right)}^{-2}={\left(\frac{3}{2}\right)}^2=\frac{9}{4}$.

(d) $4^{-1/2}$. Apply the rule with $r=\frac{1}{2}$:

Notice that this answer, like all the others, is positive, even though the exponent was negative and fractional. As the section puts it, the minus sign lives in the exponent and dictates division, not the sign of the value.

(a) $16x^7$    (b) $3a^2$

Work from the inside out: deal with the brackets first, then combine what is left.

Part (a).

Step 1. Expand the bracket using the Power of a Product Rule, $(bc{)}^n=b^n{c}^n$. The base inside is the product $2\cdot x^3$, so the exponent $4$ goes onto both factors:

Step 2. Work out each piece. First $2^4=16$. Then use the Power of a Power Rule on the second piece, multiplying the exponents:

So the numerator is $16x^{12}$.

Step 3. Now divide by $x^5$ using the Quotient Rule, which subtracts exponents. The number $16$ is untouched, since there is no number to divide it by:

Part (b).

Step 1. Expand the bracket:

Here $3^3=27$ and the exponents $2$ and $3$ multiply to give $a^6$.

Step 2. Divide, handling the numbers and the letters separately:

Check part (b) with a number. Put $a=2$. The original expression is

and our simplified answer gives $3\cdot 2^2=3\cdot 4=12$. They agree.

The most frequent error in problems like these is forgetting to raise the number inside the bracket to the power — writing ${\left(3a^2\right)}^3=3a^6$ instead of $27a^6$. The check above would have caught it immediately.

(a) $4$    (b) $-4$    (c) ${\displaystyle \frac{3}{4}}$

These run the earlier exercises backwards: instead of being given an exponent and asked for a value, we are given the value and asked for the exponent.

(a) $3^{?}=81$. Multiply $3$'s together until $81$ appears:

So the exponent is $4$.

(b) $b^{?}={\displaystyle \frac{1}{b^4}}$. The Negative Exponent Rule says $b^{-r}={\displaystyle \frac{1}{b^r}}$. Reading it from right to left with $r=4$:

So the exponent is $-4$. The rule is what converts "one over" into a minus sign upstairs.

(c) ${16}^{?}=8$. This one needs more thought, because $8$ is smaller than $16$, so the exponent will be a fraction between $0$ and $1$.

The way in is to write both numbers as powers of the same base. Both are powers of $2$:

Let the unknown exponent be $r$. Then

using the Power of a Power Rule. We want this to equal $2^3$, so we need the exponents to match:

Dividing both sides by $4$ gives

Check with the Rational Exponent Rule: take the fourth root of $16$ first, which is $2$, then cube it to get $8$. That is exactly ${16}^{3/4}=8$, matching part (c) of an earlier exercise.

The technique used here — rewriting both sides as powers of a common base — is the standard way to attack a problem where the unknown sits in the exponent.

$\sqrt{(-5{)}^2}=5$ and $\sqrt[3]{(-5{)}^3}=-5$.

Work each one out from the inside.

$\sqrt{(-5{)}^2}$. First square:

The minus signs cancel, leaving a positive number. Now take the square root:

So $\sqrt{(-5{)}^2}=5$, not $-5$.

$\sqrt[3]{(-5{)}^3}$. First cube:

With three factors one minus sign has no partner, so the result is negative. Now take the cube root: we need $u$ with $u^3=-125$, and $u=-5$ works. So

Why they differ. The two cases are governed by the last two lines of the table of root properties in this section:

When $n$ is odd, raising to the $n$th power preserves the sign, so the root can undo it exactly and we get $a$ back unchanged.

When $n$ is even, raising to the $n$th power destroys the sign — both $5$ and $-5$ have square $25$. The root cannot know which one it started from, and by convention it returns the positive one. So we get $|a|$, not $a$.

With $a=-5$ the two formulas give $-5$ and $|-5|=5$ respectively, exactly the answers found above.

The practical rule. $\sqrt{{\text{something}}^2}$ is never negative. If you ever produce a negative answer from an even-index root, you have made a sign error.

(a) $5$ m    (b) $27$ times

Part (a). For a cube with edge $s$, the volume is $s$ multiplied by itself three times, that is,

We are told $V=125$, so we need the number whose cube is $125$. That is exactly what the cube root does:

Search for it: $4^3=64$, too small; $5^3=125$, exactly right. So

The index $3$ is odd, so this root is the only real one — there is no second answer to worry about. And in any case a length must be positive.

Check: $5^3=5\cdot 5\cdot 5=125$. ✓

Part (b). The new edge is $3s$, so the new volume is

Apply the Power of a Product Rule, which puts the exponent onto each factor:

Since the old volume was $s^3$, the new volume is $27$ times the old one.

Check with the actual numbers. The new edge is $3\cdot 5=15$ metres, so the new volume is

and $3375÷125=27$. ✓

The point worth taking away. Tripling a length does not triple the volume — it multiplies it by $3^3=27$. This is a direct consequence of the Power of a Product Rule, and it is why the answer surprises people. The same reasoning shows that tripling the edge multiplies the surface area by $3^2=9$.

$32$ in both cases; taking the root first is far easier.

The Rational Exponent Rule offers both routes:

Here $b=64$, $m=5$ and $n=6$.

Route 1: root first. Take the sixth root of $64$. We need $u$ with $u^6=64$. Trying $u=2$:

So ${64}^{1/6}=2$. Now raise to the fifth power:

Route 2: power first. Raise $64$ to the fifth power:

Now take the sixth root of that number. We need $u$ with $u^6=\mathrm{1,073,741,824}$, and the answer is $u=32$, since ${32}^6$ is indeed that number. So

Comparing the routes. Both give $32$, as the rule promises. But Route 1 never dealt with a number larger than $64$, while Route 2 required multiplying out a ten-digit number and then recognising it as a sixth power — something almost nobody could do without a calculator.

The general advice. Always take the root first when evaluating $b^{m/n}$ by hand. The root shrinks the number down to something manageable, and the power is then applied to a small quantity. The only time to prefer the other order is when the root of $b$ is not a whole number but the root of $b^m$ happens to be.

(a) $3\sqrt[3]{2}$    (b) ${\displaystyle \frac{1}{b\sqrt[3]{b^2}}}$

Part (a): $\sqrt[3]{54}$.

The number $54$ is not a perfect cube — $3^3=27$ is too small and $4^3=64$ is too big — so the answer will not be a whole number. Instead we pull out the largest perfect cube hiding inside $54$.

Look for a factor of $54$ that is a perfect cube. Since

and $27=3^3$, we have found one. Now use the root property from the table of this section, $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$, which is the Power of a Product Rule with $r=1/3$:

Since $\sqrt[3]{27}=3$,

The remaining $\sqrt[3]{2}$ cannot be simplified further, because $2$ has no perfect-cube factor other than $1$.

Check roughly: $\sqrt[3]{2}\approx 1.26$, so $3\sqrt[3]{2}\approx 3.78$. And ${3.78}^3\approx 54$. ✓

Part (b): $b^{-5/3}$.

Step 1. Deal with the minus sign using the Negative Exponent Rule:

Step 2. Convert the fractional exponent into a radical using the Rational Exponent Rule, $b^{m/n}=\sqrt[n]{b^m}$:

Step 3. Simplify the radical, exactly as in part (a), by splitting off the largest perfect cube. Since $b^5=b^3\cdot b^2$,

So

which is the form shown in this section's own example.

Check with a number. Take $b=8$. The original gives $8^{-5/3}={\displaystyle \frac{1}{8^{5/3}}}={\displaystyle \frac{1}{2^5}}={\displaystyle \frac{1}{32}}$. Our answer gives ${\displaystyle \frac{1}{8\sqrt[3]{64}}}={\displaystyle \frac{1}{8\cdot 4}}={\displaystyle \frac{1}{32}}$. They agree.

Why it is wrong. The student has taken the Power of a Product Rule,

and applied it to a sum. But that rule is about a product $bc$, not about a sum $b+c$. There is no rule in this section — or anywhere — allowing an exponent to be distributed across a $+$ sign. The list of five laws should be read carefully on this point: every one of them involves multiplication, division, or repeated powers, and none of them involves addition of the bases.

Why the rule works for products but not sums. Go back to the definition, $b^n$ means $n$ copies of $b$ multiplied. So

where the middle step just reorders the factors, which multiplication permits. Now try the same with a sum:

and there is no way to rearrange this into $b^2+c^2$, because the two brackets have to be multiplied out and that produces extra terms.

The numerical test. Put $x=3$ and $y=4$.

The correct left-hand side:

The student's right-hand side:

Since $49\ne 25$, the claimed identity is false. A single counterexample is enough to demolish a claimed identity, and this one is worth memorising.

A remark. Try a second pair of numbers to confirm the failure is not a fluke. With $x=1$ and $y=1$, the left side is $(1+1{)}^2=4$ while the right side is $1+1=2$. Again they differ. The essential thing to carry away is the negative lesson: an exponent may never be distributed over a sum.

The same warning applies to roots, since a root is just a fractional exponent. For instance $\sqrt{9+16}=\sqrt{25}=5$, while $\sqrt{9}+\sqrt{16}=3+4=7$, and these differ.

The correct value is $4$.

The correct value. Work from the inside out, which is always the safe approach:

and then

So $\sqrt{(-4{)}^2}=4$.

The error. The student's slogan, "the square root undoes the square", is only half true. Squaring throws away the sign of a number: both $4$ and $-4$ square to $16$. Once that information is gone, no operation can recover it. The square root symbol must return a single definite value, and by the convention stated in this section it returns the non-negative one.

So squaring is not reversible in the way the student assumed. It sends two different inputs to the same output, and the root cannot tell them apart afterwards.

The rule that governs this. The table of root properties in this section states

With $a=-4$ and $n=2$ this gives

matching our computation. The absolute value in the formula is precisely the record of the lost sign.

When the student's slogan is safe. It works when the index is odd, because an odd power keeps the sign. For instance

and here the root really does undo the cube exactly. It also works for even indices when the number started out non-negative: $\sqrt{4^2}=4$.

A quick self-check. An even-index root can never return a negative value. If your answer to a square root is negative, something has gone wrong.

Sometimes true; it holds exactly when $a\ge 0$.

The answer is sometimes.

A case where it holds. Take $a=3$:

A case where it fails. Take $a=-3$:

but $a=-3$. Since $3\ne -3$, the statement is false here.

Since both outcomes occur, it is sometimes true.

Exactly when it holds. This section supplies the general formula for an even index:

So the statement $\sqrt{a^2}=a$ amounts to the claim $|a|=a$. Looking back at the definition of absolute value, $|a|=a$ precisely when $a$ is positive or zero. Therefore the statement holds exactly when

and fails for every negative $a$. When we have instead $\sqrt{a^2}=-a$, which is positive.

Why this matters. Expressions such as $\sqrt{a^2}$ appear constantly, and it is tempting to cancel the root against the square automatically. That is safe only when you know the letter stands for a non-negative number. If the sign is unknown, the correct simplification is $|a|$, keeping the bars until the sign can be settled.

A contrast. For an odd index no such caution is needed:

holds for every real $a$, positive, negative or zero, because an odd power preserves sign and the odd root is unique. So the trouble is caused entirely by the even index, not by roots in general.

What goes wrong. Look at the two factors on the left separately.

The expression $\sqrt{-4}$ asks for a real number $u$ with $u^2=-4$. But squaring any real number gives a result that is zero or positive, so no such real $u$ exists. The same applies to $\sqrt{-9}$. As this section records, $\sqrt[n]{b}$ is not a real number when and $n$ is even.

So the left-hand side $\sqrt{-4}\cdot \sqrt{-9}$ is a product of two things that do not exist as real numbers. There is nothing there to multiply, and the rule has nothing to act on.

The condition attached to the rule. The table of root properties in this section is stated under the standing assumption that all the roots involved exist as real numbers. The section then adds an explicit warning for exactly this situation: when $n$ is even and both and , the quantity $\sqrt[n]{ab}$ does exist, because $ab$ comes out positive, yet $\sqrt[n]{a}$ and $\sqrt[n]{b}$ individually do not. In that case the root may not be split.

The trap this creates. The right-hand side looks perfectly harmless:

since two negatives multiply to a positive. A student who splits the root the wrong way round may therefore feel that the calculation has succeeded, when in fact the middle step never made sense.

To be clear about what is legitimate: computing $\sqrt{(-4)(-9)}$ by multiplying first and taking the root second is entirely correct and gives $6$. What is not permitted is taking the roots first and multiplying second, since the intermediate quantities are not real numbers.

The lesson. Every rule in this section carries conditions — $b\ne 0$ for the Quotient Rule, $c\ne 0$ for the Power of a Quotient Rule, existence of the roots for the radical properties. Before applying a rule, check its conditions. Here, the order in which you multiply and take roots genuinely changes whether the working is valid.

The general argument. The Power of a Power Rule says

and this section stresses that the definitions of fractional exponents were chosen precisely so that this law continues to hold when $r$ and $s$ are not whole numbers.

Apply it with $r=\frac{1}{2}$ and $s=\frac{1}{3}$:

Multiply the two fractions in the exponent, multiplying tops together and bottoms together:

Therefore

In radical language this reads $\sqrt[3]{\sqrt{b}}=\sqrt[6]{b}$: taking a square root and then a cube root is the same as taking a sixth root in one go. That is the third line of the table of root properties in this section, with $n=2$ and $m=3$.

Verification with $b=64$.

Left side, one step at a time. First the square root:

since $8^2=64$. Then the cube root of that:

since $2^3=8$. So the left side is $2$.

Right side. We need the sixth root of $64$, that is, the number whose sixth power is $64$:

so ${64}^{1/6}=2$.

Both sides equal $2$, confirming the identity in this case.

A remark on why it works. Taking a square root splits the exponent in half, and taking a cube root splits what remains into thirds. Doing both divides the original exponent by $6$ altogether — which is why the fractions multiply rather than add. Contrast this with $b^{1/2}\cdot b^{1/3}$, where the base is the same and the exponents would be added, giving $b^{5/6}$, a different quantity entirely.

$|x|+x$; this equals $0$ at $x=-2$ and $10$ at $x=5$.

Handle the two terms separately, because they behave differently.

The first term, $\sqrt{x^2}$. The index is $2$, which is even. The table of root properties in this section gives ${\left(a^n\right)}^{1/n}=|a|$ for even $n$, so

The absolute value bars must be kept, because we are told only that $x$ is some real number and we do not know its sign.

The second term, $\sqrt[3]{x^3}$. The index is $3$, which is odd. The table gives ${\left(a^n\right)}^{1/n}=a$ for odd $n$, so

with no bars needed. An odd power preserves the sign and the odd root is unique, so nothing is lost.

Putting them together.

This can be taken further by splitting into cases, using the definition of absolute value from the earlier section.

• If $x\ge 0$, then $|x|=x$, so the expression is $x+x=2x$.
• If , then $|x|=-x$, so the expression is $-x+x=0$.

So the whole expression collapses to $0$ for every negative $x$, and to $2x$ for every non-negative $x$.

Evaluation at $x=-2$. Using the case rule, $x$ is negative, so the answer is $0$. Checking directly:

and $2+(-2)=0$. ✓

Evaluation at $x=5$. Here $x$ is positive, so the answer should be $2\cdot 5=10$. Checking directly:

and $5+5=10$. ✓

The moral. Writing $\sqrt{x^2}=x$ would have given $x+x=2x$ for every $x$, and at $x=-2$ that predicts $-4$ instead of the true value $0$. The absolute value is not decoration; dropping it changes the answer.