Why Any Number to the Zero Power Equals 1

Powers grow as the exponent climbs, each step up multiplying by the base again. This section goes the other direction and asks what sits one step below the first power, where the exponent is $0$. We will work out what $a^0$ equals and why every base gives the same value.

Problem

A lab grows a crystal that triples in width every day. Today the crystal measures $3^4$ units across. Reading the days backward instead of forward, yesterday it was $3^3$, the day before $3^2$, and so on, each earlier day being the next one divided by $3$. Keep walking backward one day at a time past $3^1$. What single number does the crystal measure on the day written $3^0$?

Show a hint

Each step backward in time divides the size by $3$. Write the chain $3^4=81,\ 3^3=27,\ 3^2=9,\ 3^1=3$ and take one more step.
The step from $3^1$ down to $3^0$ divides by $3$ like every step before it. What is $3\div 3$?

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Each day backward divides the size by $3$, so read the chain downward. $$3^4=81,\quad 3^3=27,\quad 3^2=9,\quad 3^1=3$$ The next step down, from $3^1$ to $3^0$, divides by $3$ just like all the others. $$3^1\div 3 = 3\div 3 = \boxed{1}$$ The exponent dropped from $1$ to $0$, and the size landed on $1$.

Problem

Run the same backward walk on a different base to be sure it is not a fluke about $3$. A deep-sea probe doubles its stored charge every hour and holds $2^5$ units right now. Step back hour by hour, each step halving, through $2^4, 2^3, 2^2, 2^1$, and take one final step. What number does $2^0$ come out to?

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Halving $2^1=2$ is the last step you need.
Every step back divides by the base $2$, and the step from $2^1$ to $2^0$ is no exception.

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Each hour backward halves the charge, so walk the chain down. $$2^5=32,\quad 2^4=16,\quad 2^3=8,\quad 2^2=4,\quad 2^1=2$$ One more halving carries $2^1$ down to $2^0$. $$2^1\div 2 = 2\div 2 = \boxed{1}$$ A different base, the same landing. Stepping down past the first power always divides the base by itself and arrives at $1$.

The staircase is one route to $a^0$. Here is a second one, coming straight from a law you already own, the rule for dividing powers with a common base. Two independent routes landing on the same value is what makes it convincing.

Problem

A relay station sends out $7^5$ signal pulses and an identical station receives exactly $7^5$ of them, none lost. The fraction kept is $\dfrac{7^5}{7^5}$. Read it two ways. First, any nonzero quantity divided by itself is some plain number, so what is that number? Second, the quotient rule subtracts the exponents, so what power of $7$ does the fraction become? Give the plain number that both readings must agree on.

Show a hint

Anything nonzero over itself, like $\frac{19}{19}$ or $\frac{7^5}{7^5}$, is a single familiar value.
The quotient rule turns $\frac{7^5}{7^5}$ into $7^{5-5}=7^0$. The two readings name the same quantity, so set them equal.

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Read the fraction the everyday way first. Any nonzero amount divided by itself is $1$, so $\dfrac{7^5}{7^5}=1$. Now read it through the quotient rule, which subtracts exponents on a common base. $$\frac{7^5}{7^5}=7^{5-5}=7^0$$ Both readings describe the exact same fraction, so they must be equal. $$7^0 = \boxed{1}$$ The law forces the same answer the staircase already showed.

Problem

Now mix the zero power into ordinary arithmetic so the order of operations gets a say. A jukebox charges $9^0$ tokens for a song and $2^3$ tokens for a movie clip. A student buys five songs and one clip. Evaluate the total $5\times 9^0 + 2^3$.

Show a hint

Powers come before multiplication and addition. Turn $9^0$ and $2^3$ into plain numbers first.
After replacing the powers, the expression is $5\times 1 + 8$.

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Evaluate the powers before anything else. The zero power gives $9^0=1$, and $2^3=8$. $$5\times 9^0 + 2^3 = 5\times 1 + 8$$ Multiplication next, then the addition. $$5 + 8 = \boxed{13}$$ A zero power is worth exactly $1$ wherever it appears, and order of operations treats it like any other power.

Here is where most slip-ups happen. A zero power equals $1$, yes, but you have to be sure about what the exponent is actually attached to. An exponent grips only the base sitting directly beneath it, unless parentheses hand it more to hold. The next two problems are twins that look almost identical and come out completely different.

Problem

A perfume maker scales a fragrance note by the amount $6m^0$, where $m$ is some nonzero number of drops. A rival scales the same note by $(6m)^0$. Evaluate both. Enter the value of $6m^0$.

Show a hint

In $6m^0$ the exponent sits directly on $m$, so it grips $m$ alone, not the $6$ in front.
$6m^0$ means $6\times(m^0)=6\times 1$. The $6$ is just sitting there, untouched by the exponent.

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In $6m^0$ the exponent is written directly above $m$, so it grips only $m$. The $6$ is a separate factor multiplied in afterward. $$6m^0 = 6\times(m^0) = 6\times 1 = \boxed{6}$$ For contrast, $(6m)^0$ wraps the whole product inside the parentheses, so the exponent grips all of $6m$ and $(6m)^0=1$. Same symbols, but the parentheses move the answer from $6$ all the way to $1$.

Problem

Same idea, a little trickier. A scoreboard shows the quantity $-4^0$ on one panel and $(-4)^0$ on the next. They differ only by a pair of parentheses. Evaluate the first one, $-4^0$.

Show a hint

The exponent grips only the $4$. The minus sign is applied last, after the power, the same way $-4^2$ meant $-(4^2)$.
$-4^0$ means $-(4^0)=-(1)$. The parentheses in $(-4)^0$ would change which base the exponent holds.

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The exponent in $-4^0$ grips only the $4$, and the minus sign waits until after the power is done, exactly as $-4^2$ meant $-(4^2)$. $$-4^0 = -(4^0) = -(1) = \boxed{-1}$$ With the parentheses it flips. $(-4)^0$ lets the exponent grip the whole $-4$, and any nonzero base to the zero power is $1$, so $(-4)^0=1$. One little pair of parentheses, opposite signs.

So far the zero has been written in plain sight. It gets more interesting when the exponent is a little expression that happens to work out to zero. The power still collapses to $1$, but you have to do the exponent arithmetic first to notice.

Problem

A timer raises base $5$ to an exponent built from two settings, computing $5^{\,8-8}$. Evaluate it. Then say what the value would become if the second $8$ were changed to a $3$, giving $5^{\,8-3}$, so you can feel how fragile the collapse to $1$ really is. Enter the value of $5^{\,8-8}$.

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Always finish the exponent's own arithmetic before touching the base. What is $8-8$?
$5^{\,8-8}=5^0$, and a nonzero base to the zero power is one specific value.

Show the full solution

Settle the exponent first. Here $8-8=0$, so $$5^{\,8-8}=5^0=\boxed{1}.$$ Change that second $8$ to a $3$ and you get $5^{\,8-3}=5^5=3125$. The collapse to $1$ only happens when the exponent arithmetic actually lands on zero.

Problem

A puzzle box opens only if a stored power equals $1$. The power is $12^{\,k-k}$ for whatever nonzero number $k$ the user dials in. One person dials $k=37$, another dials $k=200$, a third dials $k=1$. For how many of these three dial settings does the box open?

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Work out the exponent $k-k$ before worrying about the base $12$.
$k-k=0$ for every $k$, so $12^{\,k-k}=12^0$ no matter what was dialed.

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The exponent is $k-k$, which is $0$ for every value of $k$, no matter how large or small. $$12^{\,k-k}=12^0=1$$ So the stored power is $1$ for $k=37$, for $k=200$, and for $k=1$ alike. The box opens on all $\boxed{3}$ settings. The base $12$ never mattered, only that the exponent collapsed to zero.

Two problems left, and they pull the whole section together. A zero exponent that only shows up once you force a common base, an exponent you have to compute before it collapses, and a power that grips only part of a product, with order of operations setting the order of every step.

Problem

Rewriting to a common base can hide a zero exponent until you dig it out. Simplify $\dfrac{16^3}{2^{12}}$ to a single power of $2$, then give its value as a plain number.

Show a hint

Write $16$ as a power of $2$ so the whole fraction lives on one base.
$16^3=(2^4)^3=2^{12}$. Now divide by subtracting exponents and watch where the exponent lands.

Show the full solution

Put everything on base $2$. Since $16=2^4$, the power rule gives $16^3=(2^4)^3=2^{12}$. Now the fraction shares one base, so subtract exponents. $$\frac{16^3}{2^{12}}=\frac{2^{12}}{2^{12}}=2^{12-12}=2^0=\boxed{1}$$ Once everything was on one base, the exponent collapsed to zero, and the value is $1$. Of course $\frac{2^{12}}{2^{12}}$ had to be $1$ anyway, a thing over itself.

Problem

Evaluate $$10 - 3\,w^{\,5-5} + (4w)^0$$ for any nonzero $w$. The expression is built so the value never depends on $w$ at all. Find that single value.

Show a hint

Two zero powers are hiding here. One has an exponent that computes to zero, $w^{\,5-5}$. The other wraps a whole product, $(4w)^0$.
$w^{\,5-5}=w^0=1$ but it is multiplied by $3$, so that term is $3\times 1=3$. And $(4w)^0=1$ on its own. Now it is plain arithmetic.

Show the full solution

The exponent $5-5=0$ sits on $w$ alone, so $w^{\,5-5}=w^0=1$ and $3\,w^{\,5-5}=3$. In $(4w)^0$ the parentheses hand the whole product to the exponent, so $(4w)^0=1$. Working left to right, $$10 - 3 + 1 = \boxed{8}.$$ Both zero powers collapse to $1$ whatever $w$ is, which is why $w$ never reaches the answer.

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Practice

You already ride the exponent laws up and down. This set walks one specific rung, the strange step where the exponent lands on zero, and asks what a base raised to nothing could possibly equal. Two different roads lead to the same place.

Practice

A tide chart counts grains of light on a sensor. At three flashes the reading is $6^3=216$, at two flashes $6^2=36$, at one flash $6^1=6$. Each step down the chart divides the reading by $6$. Take one more step down, to zero flashes, and read off $6^0$.

Show the solution

Each rung down the chart divides by the base. The reading at one flash is $6$, so the step down to zero flashes is $$6^0=6\div6=\boxed{1}.$$ This is not guesswork. Once each step is a division by $6$, the rung below $6^1$ has to be $1$.

Practice

A relay station sends $4^9$ pulses east and the same $4^9$ pulses west, then asks how many times as many went east as west. That is the quotient $\dfrac{4^9}{4^9}$. Find it two ways, once by the quotient rule that subtracts the exponents, and once by remembering what any number over itself must be. What single value do both give?

Show the solution

By the quotient rule, $$\dfrac{4^9}{4^9}=4^{9-9}=4^0.$$ But the same fraction is just one quantity over an identical copy of itself, which is always $1$. The two readings have to agree, so $$4^0=\boxed{1}.$$ Equal pulses east and west, so of course the ratio is one, and that pins down what $4^0$ means.

Practice

For any nonzero base $a$, $$a^0=1.$$ Two roads arrive here. Stepping down a ladder of powers divides by $a$ each time, so the rung below $a^1$ is $a\div a=1$. And the quotient rule turns $\dfrac{a^n}{a^n}$ into $a^{n-n}=a^0$, which is also $1$ because anything nonzero over itself is $1$. The zero power is exactly the value that keeps the exponent laws running with no exceptions.

Practice

A vault timer raises $3$ to an exponent it computes on the spot, and today that exponent is $7-7$. The whole reading is therefore $3^{\,7-7}$. What number does the vault display?

Show the solution

Finish the exponent first. Since $7-7=0$, the reading is $$3^{\,7-7}=3^0=\boxed{1}.$$ Whenever the exponent collapses to zero, the whole power flattens to $1$, no matter how large the base looks.

Practice

A trivia host stacks three different scores, $5^0$ for the first team, $12^0$ for the second, and $100^0$ for the third, then adds them. A player groans that the third team must dwarf the others. Find the sum and explain why the sizes of the bases never mattered.

Show the solution

A zero power erases the base entirely. Every nonzero base to the zero power is $1$, so $5^0=1$, $12^0=1$, and $100^0=1$. Adding the three equal terms, $$1+1+1=\boxed{3}.$$ The $100$ carried no extra weight. Under a zero exponent the base stops mattering, and only how many terms you have decides the total.

Practice

A florist labels two boxes. The first reads $5\times t^0$ and the second reads $(5t)^0$, where $t$ is some nonzero number of tulips. The florist swears the labels are different numbers. Find the value of each label and confirm the two are not equal.

Show the solution

In $5\times t^0$ the exponent sits on $t$ alone, so $t^0=1$ and the label is $5\times1=5$. In $(5t)^0$ the parentheses gather the whole product $5t$ under the zero, and any nonzero amount to the zero power is $1$. So the labels are $\boxed{5\text{ and }1}$, genuinely different. The florist is right. What the exponent touches is the whole story.

Practice

An exponent grips only the base written directly beneath it, unless parentheses gather more under it. So $c\times x^0=c$, because the zero touches $x$ alone and leaves the coefficient $c$ standing. But $(cx)^0=1$, because the parentheses hand the entire product to the zero. Watch the parentheses and you always know what the zero power covers.

Practice

A scoreboard evaluates $100-7\times3^0+2^4$ by the usual order of operations. What does it show?

Show the solution

Evaluate the powers first. The nonzero base gives $3^0=1$, and $2^4=16$, so the line becomes $100-7\times1+16$. Multiply next, $7\times1=7$, then work left to right. $$100-7+16=\boxed{109}$$ The zero power turned into a $1$ and slid into the arithmetic like any other number.

Practice

A thermostat shows $(-8)^0$ on its left panel and $-8^0$ on its right panel. With squares those two readings came out opposite, since $(-8)^2=64$ while $-8^2=-64$. Find each reading now and then subtract the right panel from the left.

Show the solution

On the left, the parentheses hand the whole $-8$ to the zero, and $-8$ is a nonzero base, so $(-8)^0=1$. On the right, the exponent touches only the $8$, so $-8^0=-(8^0)=-(1)=-1$. Subtracting the right panel from the left, $$1-(-1)=\boxed{2}.$$ Unlike the square case, the sizes match at $1$, but the minus on the right flips its sign, so the panels still disagree.

Practice

The zero power is not a special law bolted on from outside. It is forced by the laws you already have. Watch $a^3\times a^0$. The product rule says add exponents, giving $a^{3+0}=a^3$. So multiplying by $a^0$ must leave $a^3$ untouched, which means $a^0$ can only be the number that changes nothing under multiplication, and that number is $1$. The same argument works for any base, which is why mathematicians say $a^0=1$ is the only choice that keeps the whole system honest. 🚀

Practice

A lab writes a count as $\dfrac{49^{3}}{7^{6}}\times7^2$. Every base is really a power of $7$. Rewrite to that one base, then evaluate the whole thing.

Show the solution

Send everything to base $7$. The power rule gives $49^3=(7^2)^3=7^6$, so $$\dfrac{49^3}{7^6}\times7^2=\dfrac{7^6}{7^6}\times7^2=7^{6-6}\times7^2=7^0\times7^2.$$ The zero power is $1$, and multiplying by $1$ changes nothing, so the count is $1\times7^2=\boxed{49}$. Forcing a single base let the bulky front fraction vanish to $1$, leaving one tidy power.

Practice

A receipt shows a total of $(2^4\times5^4)\times9^0$ cents. Write the total as an ordinary number and count how many digits it has.

Show the solution

The factor $9^0$ is $1$, so it drops away and leaves $2^4\times5^4$. A $2$ and a $5$ make a $10$, and four such pairs give $$2^4\times5^4=(2\times5)^4=10^4=10000.$$ That is a $1$ trailed by four zeros, so the total has $\boxed{5}$ digits. The zero power was a harmless $1$ hiding in plain sight.

Practice

Three habits carry the zero power. When an exponent works out to zero, the whole power flattens to $1$ before anything else happens. When a power meets an identical copy in a quotient, the quotient rule sends it to a zero power, which is $1$. And a stray $a^0$ factor is just $1$ under multiplication, so it can be erased without changing a thing. Rewriting to a common base is what makes those identical copies appear.

Practice

Collapse this tower to one number. $$\dfrac{8^{5}\times2^{0}}{2^{15}}\times2^{3}$$ Send every base to $2$, and watch for the place where a power divided by its twin disappears.

Show the solution

Push everything to base $2$. The factor $2^0=1$ drops out, and $8^5=(2^3)^5=2^{15}$, so the front becomes $\dfrac{2^{15}}{2^{15}}=2^{15-15}=2^0=1$. That leaves $$1\times2^3=\boxed{8}.$$ Once every base was a $2$, the whole fraction folded down to $1$ and only $2^3$ was left.

Practice

A puzzle box first simplifies the inside of $\left(\dfrac{3^{6}}{3^{2}\times3^{4}}\right)^{25}$, then raises that result to the $25$th power. Find the number the box prints.

Show the solution

Work the inside before the outer exponent. The product rule gathers the denominator into $3^2\times3^4=3^6$, so the inside is $$\dfrac{3^6}{3^6}=3^{6-6}=3^0=1.$$ Raising that to the $25$th power, the base is now $1$, and $1$ multiplied by itself any number of times is still $1$. $$1^{25}=\boxed{1}$$ The outer exponent is large, but once the inside collapsed to $1$ the height of the tower stopped mattering.

Practice

Evaluate this whole expression, watching every parenthesis. $$(-3)^0+(-3)^2-3^0\times3^2$$ A classmate guesses it is some large number. Find the exact value.

Show the solution

Read each piece by what its exponent grips. The whole $-3$ is a nonzero base, so $(-3)^0=1$. An even power of a negative is positive, so $(-3)^2=9$. And $3^0=1$. The expression becomes $$1+9-3^0\times3^2=1+9-1\times9.$$ Order of operations multiplies before subtracting, so $1\times9=9$, leaving $$1+9-9=\boxed{1}.$$ Two different zero powers, two different jobs, and the whole thing settles on a single $1$.

The Zero Power

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