What a Negative Exponent Actually Means

Every power you have met so far makes a number bigger. Exponents do not have to stop at zero, though. Each step down the ladder divides by the base, and that pattern keeps right on going below zero. This section is about the powers you find down there.

Problem

A glacier lab logs ice thickness in powers of 5 centimetres. Reading down the column the entries are $5^3 = 125$, $5^2 = 25$, $5^1 = 5$, and each step down divides the thickness by 5. The next entry is $5^0$. What number is $5^0$?

Show a hint

Take the entry above, $5^1 = 5$, and divide it by 5 the way every step down does.
Dividing 5 by 5 leaves a single value, and it matches the rule that any nonzero base to the zero power gives the same number.

Show the full solution

Each step down divides by 5. From $5^1 = 5$, one more step gives $5 \div 5 = 1$, so $5^0 = \boxed{1}$. The zero power sits exactly where the dividing-by-5 pattern says it should.

Problem

Keep going down the same column. Below $5^0 = 1$ the lab writes $5^{-1}$, still made by dividing the entry above it by 5. What number is $5^{-1}$? Write it as a fraction.

Show a hint

The entry above is $5^0 = 1$. Divide it by 5 like every other step.
Dividing 1 by 5 gives a fraction smaller than 1. The exponent went negative and so the number dropped below 1.

Show the full solution

One more step down divides $5^0 = 1$ by 5, giving $1 \div 5 = \dfrac{1}{5}$. So $5^{-1} = \boxed{\dfrac{1}{5}}$. A negative exponent has flipped the base into the bottom of a fraction.

There is a second way to see the same fact, using the quotient rule you already know. Dividing powers of the same base subtracts the exponents. When the bottom exponent is the larger one, that subtraction gives a negative number.

Problem

A delivery drone climbs in stages, each multiplying its altitude by 4. Compute $\dfrac{4^2}{4^5}$ two ways. First as a plain fraction by writing out the factors and cancelling, then by the quotient rule which subtracts $2 - 5$. What negative power of 4 do both ways give? Give the exponent only.

Show a hint

Cancelling two 4s from top and bottom leaves three 4s in the denominator, so the fraction is $\dfrac{1}{4^3}$.
The quotient rule says $4^{2-5} = 4^{-3}$, and that must equal the $\dfrac{1}{4^3}$ you found. The exponent is the value you read off.

Show the full solution

Writing the factors, two 4s cancel and three remain underneath, so $\dfrac{4^2}{4^5} = \dfrac{1}{4^3}$. The quotient rule agrees, $4^{2-5} = 4^{-3}$. The two views match, $4^{-3} = \dfrac{1}{4^3}$, and the exponent is $\boxed{-3}$.

Problem

A pollen grain seen under a microscope has width $3^{-4}$ of a millimetre. Write $3^{-4}$ as a single fraction.

Show a hint

A negative exponent puts the matching positive power in the denominator.
So this is $\dfrac{1}{3^4}$, and $3^4 = 81$.

Show the full solution

Flip the base under the bar with its matching positive power. $$3^{-4} = \dfrac{1}{3^4} = \dfrac{1}{81}$$ So the width is $\boxed{\dfrac{1}{81}}$ of a millimetre.

A negative exponent means a reciprocal, and that single idea also runs in reverse. If the base is already a fraction, a negative exponent flips it back up. The next two problems push on that.

Problem

A zoom lens reverses its setting. The reduced view was $\dfrac{1}{4}$, and reversing it raises that to the power $-3$. Evaluate $\left(\dfrac{1}{4}\right)^{-3}$ as a whole number.

Show a hint

A negative exponent takes the reciprocal of the base. The reciprocal of $\dfrac{1}{4}$ is 4.
So $\left(\dfrac{1}{4}\right)^{-3} = 4^3$. Now cube 4.

Show the full solution

A negative exponent flips the base over, and the reciprocal of $\dfrac{1}{4}$ is 4. $$\left(\dfrac{1}{4}\right)^{-3} = 4^{3} = 64$$ So the value is $\boxed{64}$. A fraction raised to a negative power turns upright and grows.

Problem

A relay stores the reciprocal of 5 as $5^{-1}$, then a reset raises that stored value to the power $-2$. Evaluate $\left(5^{-1}\right)^{-2}$ as a whole number.

Show a hint

A power to a power multiplies the exponents, $(-1)\times(-2)$, or read $5^{-1}$ as the reciprocal $\dfrac{1}{5}$ and flip it back up.
Either way you land on $5^2$. Now square 5.

Show the full solution

A power to a power multiplies the exponents, and $(-1)\times(-2) = 2$. $$\left(5^{-1}\right)^{-2} = 5^{(-1)\times(-2)} = 5^{2} = 25$$ The value is $\boxed{25}$. Two flips undo each other, so the reciprocal comes back upright.

Here is where negative powers become genuinely useful. Once exponents can go negative, the product and quotient rules work across zero without any new rules. Adding exponents is just adding numbers, and some of those numbers are now below zero.

Problem

A cipher key is built as $7^{6} \times 7^{-4}$. Use the product rule to combine it into a single power of 7, then give that power's value.

Show a hint

Adding the exponents is allowed even when one is negative. Add $6$ and $-4$.
$6 + (-4) = 2$, so the key is $7^2$.

Show the full solution

Add the exponents. $$7^{6} \times 7^{-4} = 7^{6 + (-4)} = 7^{2} = 49$$ The value is $\boxed{49}$. Adding a negative exponent is the same as subtracting, so the product rule works straight across zero.

Problem

A camera meter sets two opposing exposure stops that multiply to $9^{5} \times 9^{-5}$. Combine them into a single power of 9 and give its value.

Show a hint

Add the exponents, $5 + (-5)$.
The exponents cancel to zero, and any nonzero base to the zero power gives the same number.

Show the full solution

The exponents are opposites and add to zero. $$9^{5} \times 9^{-5} = 9^{5 + (-5)} = 9^{0} = 1$$ A power times its reciprocal collapses to $\boxed{1}$.

Problem

A spinning rotor multiplies its energy by 4 each round. After one round the energy is $4^1$ units, but the formula divides by the rotor's idle reading $4^{-3}$, giving $\dfrac{4^1}{4^{-3}}$. Write the result as a single power of 4 and give its value.

Show a hint

Dividing by $4^{-3}$ subtracts that exponent, $1 - (-3)$. Subtracting a negative adds.
$1 - (-3) = 4$, so the result is $4^4$.

Show the full solution

The quotient rule subtracts the bottom exponent. $$\dfrac{4^{1}}{4^{-3}} = 4^{1 - (-3)} = 4^{4} = 256$$ The value is $\boxed{256}$. Subtracting a negative adds, so dividing by a negative power raises the exponent.

Negative bases and negative exponents can show up together, and then the parentheses decide the sign. The next problem lines up three expressions that look almost the same, and only one of them is positive. Check what each exponent applies to and what each minus sign covers.

Problem

Exactly one of these three is a positive number. $$-2^{-4}, \qquad (-5)^{-2}, \qquad (-2)^{-3}$$ Find the one that is positive and give its value as a fraction.

Show a hint

In $-2^{-4}$ the exponent grips only the 2 and the minus waits outside, so it is $-\dfrac{1}{16}$, a negative. In $(-5)^{-2}$ the parentheses trap the sign and an even power makes it positive.
In $(-2)^{-3}$ the sign is inside but an odd power keeps it negative, $-\dfrac{1}{8}$. That leaves $(-5)^{-2}$ as the only positive, so evaluate $\dfrac{1}{(-5)^2}$.

Show the full solution

$(-5)^{-2}$ is the positive one. The parentheses hold the sign inside, and an even power of a negative is positive, so $(-5)^{-2} = \dfrac{1}{(-5)^2} = \boxed{\dfrac{1}{25}}$. The other two miss for different reasons. In $-2^{-4}$ the exponent touches only the 2, leaving $-\dfrac{1}{16}$, and in $(-2)^{-3}$ the sign is inside but an odd power keeps it negative at $-\dfrac{1}{8}$.

Letters behave exactly like numbers here. The same exponent laws combine variable powers across zero and into the negatives. The last two problems put everything from this section together.

Problem

A growth model uses the expression $t^{3} \times t^{-7}$. Simplify it to a single power of $t$, written with one exponent. Give the exponent.

Show a hint

The product rule adds the exponents even when they are mixed in sign. Add $3$ and $-7$.
$3 + (-7) = -4$, so the result is $t^{-4}$. The exponent is what you read off.

Show the full solution

Add the exponents, sign and all. $$t^{3} \times t^{-7} = t^{3 + (-7)} = t^{-4}$$ The exponent is $\boxed{-4}$. Written without a negative, that is $\dfrac{1}{t^4}$.

Problem

Four dials each set a magnification, all powers of 3. They read $$9^{-3}, \qquad \left(\tfrac{1}{27}\right)^{2}, \qquad \left(\tfrac{1}{3}\right)^{-5}, \qquad \left(3^{2}\right)^{-4}.$$ Rewrite each as a single power of 3, then say which dial gives the smallest magnification. Give that dial's value as a power of 3, for example in the form 3^7.

Show a hint

Send each to base 3. A fraction to a negative power flips upright, a fraction to a positive power keeps the base downstairs, and a power to a power multiplies the exponents.
You should get $3^{-6}$, $3^{-6}$, $3^{5}$, and $3^{-8}$. With one base, the smallest exponent loses.

Show the full solution

Send every dial to base 3. $9^{-3} = (3^2)^{-3} = 3^{-6}$, $\left(\tfrac{1}{27}\right)^{2} = 27^{-2} = (3^3)^{-2} = 3^{-6}$, $\left(\tfrac{1}{3}\right)^{-5} = 3^{5}$, and $\left(3^{2}\right)^{-4} = 3^{2\times(-4)} = 3^{-8}$. With one base the smallest exponent gives the smallest magnification, and the smallest of $-6, -6, 5, -8$ is $-8$, so the answer is $\boxed{3^{-8}}$, which equals $\dfrac{1}{6561}$.

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Practice

You already know the exponent ladder climbs by multiplying, and that any nonzero base raised to the zero power lands on $1$. The questions below walk that same ladder the other way, below zero, where each step still divides by the base. Once you see what a negative exponent really asks for, a whole shelf of reciprocal tricks opens up.

Practice

A scientist photographs a snowflake at a magnification of $5^2$. To study a real raindrop instead, she wants the opposite, a view shrunk by that same factor, which she writes as $5^{-2}$. Continue her ladder by hand. Starting from $5^2=25$, each step down divides by $5$, so $5^1=5$ and then $5^0=1$. Take two more steps down and give the value of $5^{-2}$ as a fraction.

Show the solution

Each step down the ladder divides by $5$. $$5^2=25,\quad 5^1=5,\quad 5^0=1,\quad 5^{-1}=\tfrac{1}{5},\quad 5^{-2}=\boxed{\dfrac{1}{25}}$$ The values keep shrinking by the same factor past zero, which is why $5^{-2}$ is the same as $\dfrac{1}{5^2}$.

Practice

The ladder never breaks at zero. For any nonzero base $b$, one more step down from $b^0=1$ keeps dividing, so a negative exponent means a reciprocal. $$b^{-n}=\dfrac{1}{b^n}$$ Read $b^{-n}$ as one over $b$ raised to the matching positive power.

Practice

A relay tower's signal fades by a factor of $2$ every kilometer. At the tower the strength reads $2^3$ units. A listener sits far enough away that the strength has fallen to $2^{-4}$ units. Written as a single fraction, how strong is the signal where the listener sits?

Show the solution

A negative exponent sends the power to the denominator. $$2^{-4}=\dfrac{1}{2^4}=\boxed{\dfrac{1}{16}}$$ The $2^3$ at the start was never needed, it only sets the scene.

Practice

A jeweler grades polishing grit by powers of $7$, where a finer grit carries a smaller number. The finest cloth on the bench is labeled $7^{-3}$. Rewrite that label with a positive exponent only, then give its value as a fraction.

Show the solution

$7^{-3}$ is 1 over the matching positive power, so with a positive exponent the label reads $\dfrac{1}{7^3}$. $$7^{-3}=\dfrac{1}{7^3}=\boxed{\dfrac{1}{343}}$$

Practice

Two filters sit in a beam of light. The first passes $4^{2}$ of every batch through, the second passes $4^{-5}$ of whatever reaches it. Multiply the two factors and write the combined factor as a single power of $4$.

Show the solution

Same base, so multiplying adds the exponents and the minus sign rides along. $$4^{2}\times4^{-5}=4^{2+(-5)}=\boxed{4^{-3}}$$ The law that added exponents above zero keeps working straight across zero and into the negatives.

Practice

Every law of exponents you have met still holds when an exponent goes negative. Adding exponents, subtracting them, and multiplying them all carry the minus sign along without complaint. A negative exponent is a fully grown number on the ladder, not a special case.

Practice

A timekeeper records a tiny span as $\dfrac{10^{2}}{10^{6}}$ of a day. Use the quotient rule to write this as a single power of $10$, then say how many places after the decimal point its leading digit sits.

Show the solution

Subtract the exponents, bottom from top. $$\dfrac{10^2}{10^6}=10^{2-6}=10^{-4}$$ That value is $\dfrac{1}{10^4}=0.0001$, so the leading $1$ sits in the $\boxed{4}$th place after the decimal point. The quotient rule lands you below zero whenever the bottom exponent is the larger one.

Practice

This is exactly how scientists write down extremely small numbers. A nanometer is $10^{-9}$ of a meter, and a single water molecule is only a few of them across. The minus sign matters here, it counts how many times you divided by ten to shrink a meter down to something you could never see.

Practice

A clockmaker winds a spring $6$ full turns, recorded as $9^{6}$, then unwinds it $6$ turns, recorded as $9^{-6}$. Combine the two as a single power of $9$ and give the resulting value.

Show the solution

Same base, so add the exponents. $$9^{6}\times9^{-6}=9^{6+(-6)}=9^{0}=\boxed{1}$$ A power times its opposite lands back on $9^0=1$, so the spring ends where it began.

Practice

A photographer halves the light at each click of a dial. One full sweep dims the light to $\left(\dfrac{1}{2}\right)^{5}$ of the start. Her assistant insists the same dimming can be written as $2^{-5}$. Show they agree by giving the single fraction both equal.

Show the solution

Distribute the exponent over the fraction. $$\left(\dfrac{1}{2}\right)^5=\dfrac{1^5}{2^5}=\dfrac{1}{2^5}=2^{-5}=\dfrac{1}{32}$$ Both ways of writing it give $\boxed{\dfrac{1}{32}}$. Raising a reciprocal to the $n$ is the same thing as raising the base to the $-n$.

Practice

Flipping the base undoes the negative sign on an exponent. $$\left(\dfrac{1}{b}\right)^{n}=b^{-n}\qquad\text{and}\qquad (b^{-1})^{n}=b^{-n}$$ A reciprocal in the base and a minus in the exponent are two names for the same flip.

Practice

A vault dial is calibrated so that the setting $\dfrac{8}{8^{-3}}$ opens it. Simplify that to a single power of $8$ and give its exponent.

Show the solution

Write the top as $8^1$ and subtract the bottom exponent. $$\dfrac{8}{8^{-3}}=\dfrac{8^1}{8^{-3}}=8^{1-(-3)}=8^{4}$$ So the exponent is $\boxed{4}$. Subtracting a negative adds, so dividing by a negative power raises the exponent.

Practice

In a program, a variable holds $x^{6}$ and gets multiplied by $x^{-2}$, then the result is multiplied once more by $x^{-1}$. Write the final expression as a single power of $x$.

Show the solution

All one base, so the exponents simply add. $$x^{6}\times x^{-2}\times x^{-1}=x^{6+(-2)+(-1)}=x^{6-3}=\boxed{x^{3}}$$ The negative exponents pull the total down without changing how the rule works.

Practice

Musicians live on this ladder without naming it. Drop an octave and a note's frequency is halved, which is one step of $2^{-1}$. Drop another and you are at $2^{-2}$. The lowest pipe on a giant organ is many such steps below middle ground, a stack of reciprocals you can feel in your chest more than hear.

Practice

A careless student claims $-2^{-4}$ and $(-2)^{-4}$ are equal. Work out each value exactly, then state the distance between them on the number line.

Show the solution

Both values are reciprocals, but the parentheses change the sign. In $-2^{-4}$ the power touches only the $2$, so $$-2^{-4}=-\dfrac{1}{2^4}=-\dfrac{1}{16}.$$ In $(-2)^{-4}$ the sign stays inside, and a negative base to an even power comes out positive, so $$(-2)^{-4}=\dfrac{1}{(-2)^4}=\dfrac{1}{16}.$$ They sit on opposite sides of zero, so the distance is $$\dfrac{1}{16}+\dfrac{1}{16}=\dfrac{2}{16}=\boxed{\dfrac{1}{8}}.$$ The exponent grips exactly what it touches, even when it is negative.

Practice

Three lenses are labeled by how much they enlarge an image. Lens A is $\left(\dfrac{1}{8}\right)^{-5}$, lens B is $\left(\dfrac{1}{4}\right)^{-7}$, and lens C is $16^{3}$. Rewrite all three as powers of $2$ and name the single strongest lens.

Show the solution

Flip each reciprocal base, then send everything to base $2$. $$\left(\dfrac{1}{8}\right)^{-5}=8^{5}=(2^3)^5=2^{15}$$ $$\left(\dfrac{1}{4}\right)^{-7}=4^{7}=(2^2)^7=2^{14}$$ $$16^{3}=(2^4)^3=2^{12}$$ With one base the largest exponent wins, and $15>14>12$, so the strongest is $\boxed{\text{Lens A}}$ at $2^{15}$. Rewriting all three as powers of 2 is what makes them comparable at a glance.

Practice

Collapse this into a single power of $2$, exponent and all. $$\dfrac{(2^{-3})^{-4}\times 4^{-5}}{8^{-1}}$$

Show the solution

Send every piece to base $2$. A power of a power multiplies exponents, so $$(2^{-3})^{-4}=2^{(-3)\times(-4)}=2^{12}.$$ Also $4^{-5}=(2^2)^{-5}=2^{-10}$ and $8^{-1}=(2^3)^{-1}=2^{-3}$. The numerator is $2^{12}\times2^{-10}=2^{2}$, and dividing by $2^{-3}$ subtracts a negative, which adds. $$\dfrac{2^{2}}{2^{-3}}=2^{2-(-3)}=\boxed{2^{5}}$$ Every minus sign here rides along as an ordinary number, in each law it touches.

Negative Powers

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