# It’s the Law — the Laws of Exponents

Copyright © 2002–2017 by Stan Brown

Copyright © 2002–2017 by Stan Brown

**Summary:**
The rules for combining powers and roots seem to confuse a
lot of students. They try to memorize everything, and of course it’s
a big mishmash in their minds. But
**the laws just come down to counting**, which anyone can do, plus three
definitions to memorize. This page sorts out what you have to memorize
and what you can do based on counting, to solve every problem
involving exponents.

**See also:**
Combining Operations (Distributive Laws) includes
lots of **common mistakes** students make, with plenty of
exercises to test yourself.

There’s nothing mysterious! An exponent is simply shorthand for
**multiplying that number of identical factors**.
So 4³ is the same as (4)(4)(4), three identical factors of 4. And
x³ is just three factors of x, (x)(x)(x).

One warning: Remember the order of operations. Exponents are the
first operation (in the absence of grouping symbols like parentheses),
so the exponent applies
**only to what it’s directly attached to**.
3x³ is 3(x)(x)(x), not (3x)(3x)(3x). If we
wanted (3x)(3x)(3x), we’d need to use grouping: (3x)³.

A negative exponent means to **divide** by
that number of factors **instead of multiplying**.
So 4^{−3} is the same as
1/(4^{3}), and x^{-3} = 1/x^{3}.

As you know, you can’t divide by zero. So there’s a
restriction that x^{−n} = 1/x^{n} only when x is
not zero. When x = 0, x^{−n} is undefined.

A little later, we’ll look at negative exponents in the bottom of a fraction.

A fractional exponent—specifically, an exponent of the
form 1/n—means to **take the nth root**
instead of multiplying or dividing. For example, 4^{(1/3)} is
the 3rd root (cube root) of 4.

You can’t use counting techniques on an expression like
6^{0.1687} or 4.3^{π}. Instead, these
expressions are
evaluated using logarithms.

And that’s *it* for memory work. Period. If you memorize
these three definitions, you can work everything else out by combining
them and by counting:

Granted, there’s a little bit of hand waving in my statement that you can work everything else out. Let me make good on that promise, by showing you how all the other laws of exponents come from just the three definitions above. The idea is that you won’t need to memorize the other laws—or if you do choose to memorize them, you’ll know why they work and you’ll find them easier to memorize accurately.

1. Write 11³ as a multiplication.

2. Write j^{-7} as a fraction, using only positive
exponents.

3. What’s the value of 100^{½}?

4. Evaluate −5^{-2} and
(−5)^{-2}.

[ Answers ]

Suppose you have (x^{5})(x^{6}); how do you
simplify that? Just remember that you’re counting factors.

x^{5} = (x)(x)(x)(x)(x)
and x^{6} = (x)(x)(x)(x)(x)(x)

Now multiply them together:

(x^{5})(x^{6}) =
(x)(x)(x)(x)(x)(x)(x)(x)(x)(x)(x) = x^{11}

Why x^{11}? Well, how many x’s are there? Five x
factors from x^{5}, and six x factors from x^{6},
makes 11 x factors total. Can you see that whenever you multiply
*any* two powers of the same base, you end up with a number of
factors equal to the total of the two powers? In other words,
**when the bases are the same**, you find the
new power by just **adding the exponents**:

**Caution!** The rule above works only when multiplying powers
of the same base. For instance,

(x^{3})(y^{4}) =
(x)(x)(x)(y)(y)(y)(y)

If you write out the powers, you see there’s no way you can combine them.

Except in one case:
**If the bases are different but the exponents are the same**,
then you can **combine them**. Example:

(x³)(y³) = (x)(x)(x)(y)(y)(y)

But you know that it doesn’t matter what order you do your multiplications in or how you group them. Therefore,

(x)(x)(x)(y)(y)(y) = (x)(y)(x)(y)(x)(y) = (xy)(xy)(xy)

But from the very definition of powers, you know that’s the same as (xy)³. And it works for any common power of two different bases:

It should go without saying, but I’ll say it anyway: all the laws of exponents work in both directions. If you see (4x)³ you can decompose it to (4³)(x³), and if you see (4³)(x³) you can combine it as (4x)³.

What about dividing? Remember that dividing is just multiplying
by 1-over-something. So all the laws of division are really just laws
of multiplication. The extra definition of x^{-n} as
1/x^{n} comes into play here.

Example: What is x^{8}÷x^{6}? Well, there are
several ways to work it out. One way is to say that
x^{8}÷x^{6} =
x^{8}(1/x^{6}), but using the definition of
negative
exponents that’s just x^{8}(x^{-6}). Now use the
product rule (two powers of the same
base) to rewrite it as
x^{8+(-6)}, or x^{8-6}, or x^{2}. Another
method is simply to go back to the definition:
x^{8}÷x^{6} =
(xxxxxxxx)÷(xxxxxx) = (xx)(xxxxxx)÷(xxxxxx) =
(xx)(xxxxxx÷xxxxxx) = (xx)(1) = x^{2}. However
you slice it, you come to the same answer:
**for division with like bases you subtract exponents**,
just as for multiplication of like
bases you add exponents:

But there’s no need to memorize a special rule for division: you can always work it out from the other rules or by counting.

In the same way, **dividing different bases**
can’t be simplified unless the exponents are equal.
x³÷y² can’t be combined because it’s just xxx/yy; But
x³÷y³ is xxx/(yyy), which is (x/y)(x/y)(x/y), which is
(x/y)³.

Multiplication and division have equal
precedence, so xxx/yyy would literally mean x, times x, times x,
divided by y, times y, times y and would be equal to xxxy.
That’s why the parentheses are necessary: xxx/(yyy), as reader
Chase Ries pointed out. I had written xxx/yyy, because we often omit
the parentheses in a fraction that
doesn’t contain additions or
subtractions. But it’s best not to force the reader to puzzle out
from the context whether some parentheses have been omitted.

Parentheses around the xxx — (xxx)/(yyy) — would not be wrong, but they’re not needed because the standard order of operations is to multiply x by x by x, with or without parentheses.

As that example illustrates, you can combine like exponents even when the bases are different:

What about dividing by a negative power, like
y^{5}/x^{−4}? Use the rule you already know for
dividing:

5 5 5 4 5 4 y y y x y x 4 5 --- = -------- = -------- * -- = ----- = x y -4 ( 4) ( 4) 4 x (1 / x ) (1 / x ) x 1

But that’s much too elaborate.
Since 1 / (1/x) is just x, a negative exponent just moves
its power to the **other side of the fraction bar**.
So x^{−4} = 1/(x^{4}),
and 1/(x^{−4}) = x^{4}.

Because this article helps you,

please click to donate!Because this article helps you,

please donate at

BrownMath.com/donate.

please click to donate!Because this article helps you,

please donate at

BrownMath.com/donate.

Write each of these as a single positive power. (I’ve slipped in one or two that can’t be simplified, just to keep you on your toes.)

5. a^{7} ÷ b^{7}

6. 11² × 2³

7. 8³ x³

8. 5^{4} × 5^{6}

9. p^{11} ÷ p^{6}

10. r^{-11} ÷ r^{-2}

[ Answers ]

What do you do with an expression like (x^{5})^{4}?
There’s no need to guess—work it out by counting.

(x^{5})^{4} =
(x^{5})(x^{5})(x^{5})(x^{5})

Write this as an array:

x^{5} = (x) (x) (x) (x) (x)

x^{5} = (x) (x) (x) (x) (x)

x^{5} = (x) (x) (x) (x) (x)

x^{5} = (x) (x) (x) (x) (x)

How many factors of x are there? You see that
there are 5 factors in each row from x^{5} and 4 rows from
( )^{4}, in all 5×4=20 factors. Therefore,

(x^{5})^{4} = x^{20}

As you might expect, this applies to any
**power of a power: you multiply the exponents**. For
instance, (k^{-3})^{-2} =
k^{(-3)(-2)} = k^{6}. In general,

I can just hear you asking, “So when do I add exponents
and when do I multiply exponents?” Don’t try to remember a
rule—work it out! When you have a
**power of a power**, you’ll always have a
rectangular array of factors, like the example above. Remember the old
rule of length×width, so the combined exponent is formed by
multiplying. On the other hand, when you’re only multiplying two powers together, like
g^{2}g^{3}, that’s just the same as stringing factors
together,

g^{2}g^{3} = (gg)(ggg) =
(ggggg) = g^{5}

You can always refresh your memory by counting simple cases, like

x^{2}x^{3} = (xx)(xxx) =
x^{5}

versus

(x^{2})^{3} = (xx)^{3} = (xx)(xx)(xx) =
x^{6}

Perform the operations to remove parentheses:

11. (x^{4})^{-5}

12. (5x²)³

[ Answers ]

You probably know that **anything to the 0 power**
is 1. But now you can see why. Consider x^{0}.

By the division rule, you know that
x^{3}/x^{3} = x^{(3−3)} = x^{0}.
But anything divided by itself is 1, so
x^{3}/x^{3} = 1. Things that are equal to
the same thing are equal to each other: if x^{3}/x^{3}
is equal to both 1 and x^{0}, then 1 must equal x^{0}.
Symbolically,

x^{0} = x^{(3−3)} =
x^{3}/x^{3} = 1

There’s one restriction. You saw that we had to
create a fraction to figure out x^{0}. But division by 0 is
not allowed, so our evaluation works for anything to the 0 power
*except* zero itself:

Evaluating 0^{0} is a topic for your calculus course.

What is the value of each of these?

13. (a^{6}b^{8}c^{10} / a^{5}b^{6}d^{7})^{0}

14. 17x^{0}

[ Answers ]

The laws of radicals are traditionally taught separately from the
laws of exponents, and frankly I’ve never understood why. A
**radical is simply a fractional exponent**:
the square (2nd) root of x is just x^{1/2}, the cube (3rd)
root is just x^{1/3}, and so on. With this fact at your
disposal, you’re in good shape.

Example: . That’s easy to evaluate! You know that the cube
(3rd) root of x is x^{1/3} and the square root of that is
(x^{1/3})^{1/2}. Then use the
power-of-a-power rule to evaluate that as
x^{(1/3)(1/2)} = x^{(1/6)}, which is the 6th root of x.

Example: . Why? Because the square root is the 1/2 power,
and the product rule for the same
power of different bases tells you
that (x^{1/2})(y^{1/2}) = (xy)^{1/2}.

So far we’ve looked at fractional exponents only where the top
number was 1. How do you interpret x^{2/3}, for instance?
Can you see how to use the power rule? Since
2/3 = (2)(1/3), you can rewrite x^{2/3} =
x^{(2)(1/3)} = (x^{2})^{1/3}, which is
. It works the other way, too: 2/3 =
(1/3)(2), so x^{2/3} =
x^{(1/3)(2)} =
(x^{1/3})^{2} =
. These are examples of the general rule:

When a power and a root are involved, the top part of the fractional exponent is the power and the bottom part is the root.

Suppose p and r are the same? Then you have, for instance,
. But that’s the same as x^{5/5}, and
5/5=1, so it’s the same as x^{1} or just x.

15. Write √x^{5} as a single power.

16. Simplify ³√(a^{6}b^{9})
(That’s the cube root or third root of a^{6}b^{9}.)

17. Find the numerical value of 27^{4/3} without using a
calculator.

[ Answers ]

Well, there you are: the laws of exponents and radicals
demystified! Just remember the
three basic definitions. When you’re not sure
about a rule, like the product rule, don’t try to remember it, just
**work it out by counting** and you’ll do just fine.

From What Is an Exponent, Anyway? —
1. 11×11×11
2. 1/j^{7}
3. √100 = 10
4. −5^{-2} = −1/25
and (−5)^{-2} = 1/25 (Excel returns 1/25
or 0.04 for both of these, but
that’s
wrong.)

From Multiplying and Dividing Powers —
5. (a/b)^{7}
6. *cannot be simplified*
as a power
expression. Numerically it’s 121×8 =
968.
7. (8x)³
8. 5^{10}
(*not* 25^{10}!)
9. p^{5}
10. r^{-11-(-2)} = r^{-9} =
1/r^{9}

From Powers of Powers —
11. x^{-20} or 1/x^{20}
12. Use the rule for powers of
different bases to start with: 5^{3}(x^{2})^{3}.
Then apply the power-of-a-power rule to get
5^{3}x^{6} or 125x^{6}

From The Zero Exponent — 13. 1 14. 17×1 = 17

From Radicals —
15. x^{5/2}
16. (a^{6}b^{9})^{1/3} =
a²b³
17. 27^{4/3} = (27^{1/3})^{4} by the
power-of-a-power law. 27^{1/3} is
the same as the cube root of 27, which is 3.
(27^{1/3})^{4} =
3^{4} = 81

**24 Mar 2017**: Thanks to Chase Ries, corrected xxx/yyy to xxx/(yyy). I added an explanation of the issue involved.- (intervening changes suppressed)
- 24 Feb 2002: New article.

Because this article helps you,

please click to donate!Because this article helps you,

please donate at

BrownMath.com/donate.

please click to donate!Because this article helps you,

please donate at

BrownMath.com/donate.

Updates and new info: https://BrownMath.com/alge/