Laws of Exponents: Free Response

The first power is three factors of $7$ and the second is five of them. Multiplying sets the two runs side by side, and since multiplication puts nothing between them, the inner grouping can be dropped:

$7^3 \times 7^5 = (7 \times 7 \times 7) \times (7 \times 7 \times 7 \times 7 \times 7).$

Now count what is in the row. Three factors came from the first group and five from the second, so eight factors of $7$ stand there in all, and eight factors of $7$ is exactly what $7^8$ means:

$7^3 \times 7^5 = 7^{3+5} = 7^8.$

The only thing that changed is how many factors there are. Every one of them is still a $7$, which is why the answer is a power of $7$ and not of anything else.

The top is seven factors of $11$ and the bottom is three:

$\frac{11^7}{11^3} = \frac{11 \times 11 \times 11 \times 11 \times 11 \times 11 \times 11}{11 \times 11 \times 11}.$

Each factor below cancels one factor above, because $\tfrac{11}{11} = 1$. There are three below, so three of the seven above are cancelled and $7 - 3 = 4$ of them survive:

$\frac{11^7}{11^3} = 11 \times 11 \times 11 \times 11 = 11^4.$

Cancelling is subtraction done to a count. The bottom removes as many factors as it has, which is why the exponent of the answer is the top count with the bottom count taken away, and why the base is untouched by the whole operation.

Go back to what the symbols stand for. A power is a pile of equal factors, and multiplying two such piles pours them into one pile. Pouring changes how many things are in the pile; it does not change what the things are. So the count of factors grows, while the factor itself, the base, is carried through untouched.

The smallest product of two powers of $7$ is one factor times one factor:

$7^1 \times 7^1 = 7 \times 7 = 49 = 7^2.$

The counting rule predicts $7^2$, and $7^2$ really is $49$, so the arithmetic agrees with it. The classmate's rule predicts a base of $49$ with the same exponent, which is

$49^2 = 2{,}401,$

not $49$. One small case is enough to show the two rules disagree and which of them the arithmetic sides with.

It is worth seeing why the classmate's rule overshoots by so much. Turning the base $7$ into $49$ is itself squaring, since $49 = 7^2$, so every factor in the pile is quietly doubled up. That is a second multiplication nobody asked for, and it is exactly what makes $49^2 = 7^4$ instead of $7^2$.

Side by side. Three factors of $6$ next to five factors of $6$ make one run of eight, so the counts add:

$6^3 \times 6^5 = 6^{3+5} = 6^8.$

There is one group here, formed by pouring two runs together.

One inside the other. The outer exponent says to use $6^3$ as a factor five times over, so there are five groups and each group holds three factors of $6$:

$(6^3)^5 = 6^3 \times 6^3 \times 6^3 \times 6^3 \times 6^3.$

Five groups of three factors is fifteen factors in all, and repeated addition of equal groups is multiplication:

$(6^3)^5 = 6^{3 \times 5} = 6^{15}.$

So the same two numbers meet an addition in one setup and a multiplication in the other, $3 + 5 = 8$ against $3 \times 5 = 15$, and the second answer carries almost twice as many factors as the first.

Work outward from the innermost bracket, as the order of operations directs. On top sits a power of a power, four groups of five factors, so the exponents multiply:

$(2^5)^4 = 2^{5 \times 4} = 2^{20}.$

Underneath sit two powers of the same base side by side, so their counts add:

$2^3 \times 2^9 = 2^{3+9} = 2^{12}.$

The expression is now one power over another with the same base and the larger count on top, so cancel in pairs, which subtracts the counts:

$\frac{2^{20}}{2^{12}} = 2^{20-12} = 2^8.$

Each step used exactly one law, and which law it was depended only on the shape the expression had at that moment. That is the whole method: look at the shape, apply the matching rule, look again.

Ravi is right that the base does not move, and that is the only thing the two setups share.

What each asks to be multiplied. In $6^3 \times 6^5$ the two powers are separate quantities being multiplied together, so their factors simply pool: three copies of $6$ joining five copies of $6$ give a run of eight, and the counts add. In $(6^3)^5$ there is only one power, and the outer exponent is an instruction about it: use the whole of $6^3$ as a factor five times. That builds five equal groups of three factors, and equal groups multiply:

$6^{3+5} = 6^8 \qquad \text{against} \qquad 6^{3 \times 5} = 6^{15}.$

Seven extra factors of $6$ is not a small discrepancy. Since $6^{15} = 6^8 \times 6^7$, the second result is $6^7$ times the first, so these are nowhere near each other.

A test for any expression. Look at where the second exponent is written. If it sits on a base of its own, with a multiplication sign between the two powers, the powers are separate quantities, and when they share a base their exponents add; when the bases differ, the product rule does not apply at all and nothing is added. If the second exponent sits outside a bracket that already contains a power, it is being applied to that whole power, and the exponents multiply, whatever the base happens to be. One reliable phrasing: ask whether the second exponent is counting copies of the base, or counting copies of the first power.

The grid has $2^7$ rows, and each row holds $2^7$ pixels, so the total is a product of two powers of the same base. That is the product rule, and the counts add:

$2^7 \times 2^7 = 2^{7+7} = 2^{14} \text{ pixels}.$

Multiplying it out, $2^{14} = 16{,}384$ pixels.

The exponent is not counting pixels. It counts factors of $2$: $2^{14}$ means fourteen $2$'s multiplied together, and the number that product comes to is the pixel count. Keeping that straight is what lets the rest of the question stay in powers instead of in five-digit numbers.

The same total can be reached a second way, because the picture is square. Its side is $2^7$ pixels and that side is used twice, which is a power of a power:

$(2^7)^2 = 2^{7 \times 2} = 2^{14} \text{ pixels}.$

Adding $7 + 7$ and multiplying $7 \times 2$ agree here for the same reason that doubling a number and adding it to itself agree.

Across one side, the question is how many runs of $2^3$ pixels fit inside $2^7$ pixels, which is a division of two powers of $2$ with the larger count on top:

$\frac{2^7}{2^3} = 2^{7-3} = 2^4 = 16 \text{ tiles across}.$

The tiles are laid in a square grid, $2^4$ of them across and $2^4$ of them down, so the number of tiles is a product of powers of the same base:

$2^4 \times 2^4 = 2^{4+4} = 2^8 = 256 \text{ tiles}.$

It is worth pausing on what each number counts. $2^4$ is tiles across a side, $2^8$ is tiles over the whole picture, and neither is a count of pixels. One tile carries $2^3 \times 2^3 = 2^6 = 64$ pixels of its own, so $256$ tiles of $64$ pixels each is

$2^6 \times 2^8 = 2^{14} \text{ pixels},$

which is the picture's own pixel count. The two routes agree, which is a real check on both.

Doubling a side multiplies its pixel count by $2$, and $2$ is $2^1$, so the product rule gives the new side:

$2 \times 2^7 = 2^1 \times 2^7 = 2^8 \text{ pixels}.$

The enlarged picture is $2^8$ across and $2^8$ down, so its pixel count is

$2^8 \times 2^8 = 2^{16} \text{ pixels}.$

Compare that with the old count by dividing, which subtracts the counts of factors:

$\frac{2^{16}}{2^{14}} = 2^{16-14} = 2^2 = 4.$

Four times as many pixels, not twice as many. The tiles behave the same way: each enlarged side holds $\tfrac{2^8}{2^3} = 2^5 = 32$ tiles, so the picture needs $2^5 \times 2^5 = 2^{10} = 1{,}024$ tiles, and $\tfrac{2^{10}}{2^8} = 2^2 = 4$ again.

The reason is that the doubling is applied twice, once in each direction. Each doubling contributes one extra factor of $2$ to the total, and two extra factors of $2$ are multiplied together, not added, so they come to $2^2 = 4$. Any count of equal cells filling the whole rectangle grows this way: stretch both directions by the same amount and the count grows by that amount used twice. Not everything on a grid does, though. Count the lines of the grid, or the corner points where they cross, and the pattern is different, because those are counted along the directions rather than over the region.

Line (i). The exponents were handled correctly, $4 + 3 = 7$, but the bases were multiplied too, turning $9$ into $81$. Multiplying $9^4$ by $9^3$ lays four factors of $9$ beside three more, and every factor in that row is still a $9$, so the product rule keeps the base:

$9^4 \times 9^3 = 9^{4+3} = 9^7.$

The two answers are far apart. Since $81 = 9^2$, the written answer is $81^7 = (9^2)^7 = 9^{14}$, a power carrying twice as many factors of $9$ as the truth.

Line (ii). Here the base survived but the wrong operation was done: the exponents were added, $3 + 2 = 5$. This setup is a power of a power, which asks for $4^3$ as a factor twice, that is two groups of three factors, so the power rule multiplies:

$(4^3)^2 = 4^{3 \times 2} = 4^6.$

Adding belongs to the other setup, where two powers stand side by side with a multiplication sign between them.

Line (iii). The exponents were divided, $10 \div 2 = 5$, presumably because the expression itself is a division. But dividing powers cancels factors in pairs, and cancelling removes as many factors as the bottom has, so the counts subtract:

$\frac{8^{10}}{8^2} = 8^{10-2} = 8^8.$

A count with two taken away is not a count cut in half, and the ten factors of $8$ on top lose exactly the two that the bottom cancels.

A rule announced for every base and every pair of exponents is brought down by one honest case, so choose numbers small enough to evaluate exactly. Take $a = 4$, $m = 2$ and $n = 3$. The left side is a sum of two ordinary numbers:

$4^2 + 4^3 = 16 + 64 = 80.$

The right side is a single power:

$4^{2+3} = 4^5 = 1{,}024.$

Those are not equal, and they miss by a wide margin rather than by a slip, so the rule is false as stated.

Why it fails is worth saying, because the failure is structural rather than accidental. Every law in this lesson comes from counting factors inside a product, and a sum is not a product. In $4^2 + 4^3$ the two powers are held apart by an addition sign; there is no single run of factors to count, and nothing for an exponent to record. The left side is the number $80$, and $80$ is where it stops.

The same test settles any proposed rule of this kind. Pick the smallest numbers that make both sides computable, and compare.

Take the laws in turn rather than accepting or rejecting the claim whole.

Where the student is right. The product rule and the quotient rule really do require a shared base. Both work by counting copies of one repeated factor, and copies of $3$ cannot be pooled with copies of $5$ into a single count, because they are not copies of the same thing. So neither of those two rules touches this expression.

Where the claim breaks. Those are not the only laws in the lesson. The power of a product says

$(a \times b)^n = a^n \times b^n,$

and a true equation can be read from right to left as well as from left to right. Read backwards it gathers two powers that share an exponent, whatever their bases:

$3^4 \times 5^4 = (3 \times 5)^4 = 15^4.$

The lesson used this law in the forward direction to split $6^3$ into $2^3 \times 3^3$; running the same line the other way puts such a product back together. As a check, $3^4 \times 5^4 = 81 \times 625 = 50{,}625$, and $15^4 = 225 \times 225 = 50{,}625$.

The second expression. In $3^4 \times 5^3$ the bases differ, so the product rule is out, and the exponents differ too, so there is no shared exponent for the power of a product to collect as they stand. Nor can one be made: pulling a common exponent out of both would need a number above $1$ that divides both $4$ and $3$, and there is none. The two powers cannot be gathered into a single power.

That is not the same as saying nothing at all can be done, and it is worth being exact about the difference. The surplus factor can be peeled off, since $3^4$ is one $3$ standing beside three more, and the two powers that remain do share an exponent:

$3^4 \times 5^3 = 3 \times 3^3 \times 5^3 = 3 \times 15^3.$

As a check, $3^4 \times 5^3 = 81 \times 125 = 10{,}125$, and $3 \times 15^3 = 3 \times 3{,}375 = 10{,}125$. The result is a number multiplied by a power, not one power, so the expression really does resist being written as a single power. The reason simply has to be stated as it is, and not as "no law applies".

So the student's sentence needs its condition named rather than its base: two powers combine under the product rule when they share a base, and a product of powers gathers into one power when they share an exponent.

The base of the outer power is the product $2^3 \times 5$, so the power of a product sends that exponent onto each factor:

$(2^3 \times 5)^4 = (2^3)^4 \times 5^4.$

The first piece is now a power of a power, and the power rule multiplies its counts:

$(2^3)^4 = 2^{3 \times 4} = 2^{12}.$

The second piece is $5$, which is $5^1$, so raising it to the fourth simply gives $5^4$; there was nothing inside it to resolve. Putting the pieces back together,

$(2^3 \times 5)^4 = 2^{12} \times 5^4.$

Notice that the two exponents come out different even though one exponent was applied to the whole base. The $2$ walked in carrying three factors of its own and the $5$ walked in carrying one, and the outer exponent multiplies whatever each factor brought with it.

Inside the bracket first. The order of operations settles the bracket before the exponent acts, so $2 \times 7 = 14$ and then

$14^3 = 14 \times 14 \times 14 = 196 \times 14 = 2{,}744.$

Exponent onto each factor. The power of a product sends the $3$ to both pieces, and each piece is then evaluated on its own:

$(2 \times 7)^3 = 2^3 \times 7^3 = 8 \times 343 = 2{,}744.$

The two routes agree, as they had to. Both are multiplying the same six factors, three $2$'s and three $7$'s; the first route keeps them interleaved as $(2 \times 7)$ three times over, while the second gathers the $2$'s together and the $7$'s together first. Reordering a product never changes its value, which is precisely the property the law is built on.

Start from the definition rather than from a rule. The base is the product $a \times b \times c$, and the exponent $3$ asks for three copies of it multiplied together:

$(a \times b \times c)^3 = (a \times b \times c) \times (a \times b \times c) \times (a \times b \times c).$

Nine factors stand in that row: three $a$'s, three $b$'s and three $c$'s, in the order the copies produced them. Multiplication may be reordered, which is the commutative property, and regrouped, which is the associative property, and between them those two properties are the whole licence to gather the like letters without changing the value:

$= (a \times a \times a) \times (b \times b \times b) \times (c \times c \times c) = a^3 \times b^3 \times c^3.$

Nothing in that argument depended on the number three, in either role. Take any positive whole-number exponent $n$, the only kind a power has been given a meaning for so far. With $n$ copies of a product of any number of pieces, each piece still turns up once per copy, so each piece ends with $n$ of itself:

$(a \times b \times c)^n = a^n \times b^n \times c^n.$

What would defeat the argument is changing the operation inside the bracket. Every move above slid factors past one another, and only multiplication permits that. Put a plus sign in the bracket and there are no separate runs of $a$'s and $b$'s to gather at all: $(2 + 5)^2$ is $7^2 = 49$, while $2^2 + 5^2$ is $4 + 25 = 29$. An exponent spreads across a product, and across a quotient for the same reason as long as the denominator is not zero, but not across a sum.