# Exponential identities

There are several identities that I haven't discussed yet involving exponentiation. The basic ones are listed in the textbook on page 296. You can use these to simplify monomials, polynomials, and even some slightly more general expressions involving division.

Each of the identities involving exponentiation has a corresponding identity involving multiplication. As exponentiation is repeated multiplication and multiplication is repeated addition, so exponential identities that also involve multiplication generally correspond to multiplicative identities that also involve addition. However, this rule doesn't apply to operations in the exponent (which is not multiplied but instead counts the multiplications); these operations are the same as in the multiplicative identity. Below I list the exponential identities together with their corresponding multiplicative identities. (Please keep in mind that I point out this correspondence only because it might help you to remember the new identities. If it's more confusing than it's worth, then forget about it; just learn the new identities for themselves.)

Exponential identityMultiplicative identity
a0 ≡ 1; 0a ≡ 0;
aman ≡ am+n for a ≠ 0 or m, n ≥ 0; ab + ac ≡ a(b + c);
a1 ≡ a; 1a ≡ a;
(am)n ≡ amn for a ≠ 0 or m ≥ 0; (ab)c ≡ a(bc);
1n ≡ 1; 0a ≡ 0;
(ab)n ≡ anbn; (a + b)c ≡ ac + bc;
a−1 ≡ 1/a; (−1)a ≡ −a.
In these identities, a, b, and c stand (as usual) for real numbers, but m and n stand only for integers, because only then have I defined exponentiation. (In Intermediate Algebra, you'll learn how to define exponentiation for rational exponents, but then some of these identities require extra conditions! Conversely, if you allow only whole numbers for m and n, then you don't need any conditions at all.)

## Monomial examples

These identities are particularly useful for simplifying monomials.

For example, suppose that you want to simplify x3 · x4. You can do this the long way, using the definition of exponentiation as repeated multiplication, and it's probably a good idea to try this a few times until you get used to the exponential identities. But you can also do this the quick way using the second identity above:

• x3 · x4 ≡ (x · x · x) (x · x · x · x) ≡ x · x · x · x · x · x · x ≡ x7;
• x3 · x4 ≡ x3+4 ≡ x7.

Similarly, suppose that you have (x3)4. Again, you can do this the long way or the short way:

• (x3)4 ≡ (x · x · x) (x · x · x) (x · x · x) (x · x · x) ≡ x · x · x · x · x · x · x · x · x · x · x · x ≡ x12;
• (x3)4 ≡ x3·4 ≡ x12.

Now consider (xy)4:

• (xy)4 ≡ (xy) (xy) (xy) (xy) ≡ (x · x · x · x) (y · y · y · y) ≡ x4y4;
• (xy)4 ≡ x4y4.

There are a few tricky situations where an exponent doesn't seem to appear. In this case, use the identity a ≡ a1. For example, to simplify x · x4:

• x · x4 ≡ x (x · x · x · x) ≡ x · x · x · x · x ≡ x5;
• x · x4 ≡ x1 · x4 ≡ x1+4 ≡ x5.
Now the short way has an extra step, using x ≡ x1.

An extreme version of this is to simplify x · x:

• x · x ≡ x2;
• x · x ≡ x1 · x1 ≡ x1+1 ≡ x2.
Now the method using identities isn't really short at all! But it's nice to see that it still works.

## Examples with division

So far, I haven't used the last identity, the one involving the reciprocal of a. Since division is multiplication by a reciprocal, this identity can be used in examples involving division.

For example, consider the expression x3/x4, that is x3 divided by x4. Since division is multiplication by a reciprocal and (using the last identity above) the reciprocal of a number is the same as that number raised to the power of −1, I can rewrite this as x3(x4)−1, which I can then simplify using the other identities. Thus, I get:

• x3/x4 — original expression;
• x3(x4)−1 — rewritten without division;
• x3x−4 — since −1 · 4 = −4;
• x−1 — since 3 − 4 = −1;
• 1/x — rewriting without negative exponents.
The last step is not really necessary; both x−1 and 1/x are perfectly good simplified expressions. Unless you get a problem that specifically asks for one form or another, then it's up to you whether to use division or negative exponents in a final answer.

There is one tricky bit to watch out for in a problem like this one. The two most commonly used exponential identities (both of them used here: the one where you multiply the exponents and the one where you add the exponents) have conditions on their usage. When working with monomials, these conditions are always satisfied, since the exponents in a monomial are always whole numbers. Here, however, the conditions fail unless x ≠ 0. So I need to check and make sure that all the expressions above are really equivalent even when x := 0. In this case, they are; all of the expressions are undefined, one way or another, when x := 0. So there's nothing to worry about in this case.

Now consider x4/x3. This is very similar to the previous example, but now I will need to worry about the possibility that x := 0. Watch as I go through almost the same reasoning:

• x4/x3 — original expression;
• x4(x3)−1 — rewritten without division;
• x4x−3 — since −1 · 3 = −3;
• x1 for x ≠ 0 — since 4 − 3 = 1, and x−3 is undefined when x := 0;
• x for x ≠ 0 — since raising to the power of 1 has no effect.
Here I had to add the clause ‘for x ≠ 0’, because x1 (or simply x, for that matter) is perfectly well defined when x := 0; it comes out to 0. Yet the original expression x4/x3 involved division by 0 when x := 0, and this is undefined; the other expressions similarly involve 0 raised to a negative power, which is also undefined. So once the last negative exponent on x disappears, I need to put in this restriction on x; otherwise the expressions won't quite be equivalent.

Here's a more complicated example showing in more detail how to keep track of these things:

• x2y4/x4y2 — original expression (for this new example);
• x2y4x−4y−2 — rewritten without division;
• x2x−4y4y−2 — rearranging multiplication;
• x−2y2 for y ≠ 0 — since 2 − 4 = −2, 4 − 2 = 2, and y−4 is undefined when y := 0;
• y2/x2 for y ≠ 0 — rewritten without negative exponents.
Notice that here I have to add the condition that y ≠ 0, and I put this in as soon as the last negative exponent on y went away; but I never needed to say that x ≠ 0, because every expression has a negative exponent on x (or else divides by x raised to a positive exponent).

Finally, I should tell you how all of the examples in this section can be easily done in a single step. The basic idea is this: Division of a power corresponds to multiplication with the opposite exponent. Since multiplying powers involves adding the exponents, this means that dividing powers involves subtracting the exponents. So in the first example, x3/x4, you simply calculuate 3 − 4 = −1 to get the answer x−1, which you can rewrite as 1/x if you wish. In the next example, x4/x3, you calculate 4 − 3 = 1; noticing that the result should be undefined when x := 0, your answer is x for x ≠ 0. Finally, in the example x2y4/x4y2, you calculate 2 − 4 = −2 for the exponent on x, 4 − 2 = 2 for the exponent on y, and notice that the result should be undefined even when y := 0, to get the result x−2y2 for y ≠ 0, which you can rewrite as y2/x2 for y ≠ 0 if you wish.

## Simplifying monomials

Just as any linear expression may be simplified to a sum of a few terms (generally one for each variable, plus a constant term), so any monomial may be simplified to a product of a few factors (again one for each variable, multiplied by a constant factor). As with simplifying linear expressions, the key to dealing with a complicated problem is to move slowly from the inside out.

For example, consider the monomial 3[2(xy)4x]3y2. This is pretty busy, but I can handle it step by step:

• 3[2(xy)4x]3y2 — original expression;
• 3[2x4y4x]3y2 — distribute the exponent 4 to the factors x and y;
• 3[2x5y4]3y2 — combine the factors x4 and x into x5 (remember that x ≡ x1);
• 3[2]3[x5]3[y4]3y2 — distribute the exponent 3 to the factors 2, x5, and y4;
• 3[8]x15y12y2 — work out 23 and multiply exponents to simplify powers of powers;
• 24x15y14 — combine the constant factors and the y factors.
So in summary, 3[2(xy)4x]3y2 ≡ 24x15y14, which has bigger constants but a simpler structure.

You can apply these same ideas to examples with division or negative exponents, as long as you watch out for division by zero (or raising zero to a negative exonent). For example, consider the monomial above, now with negative exponents instead of positive ones: 3[2(xy)−4x]−3y−2. Because of the negative exponents, this expression is not a monomial, but I can simplify it anyway:

• 3[2(xy)−4x]−3y−2 — original expression;
• 3[2x−4y−4x]−3y−2 — distribute the exponent −4 to the factors x and y;
• 3[2x−3y−4]−3y−2 — combine the factors x−4 and x into x−3 (remember that −4 + 1 = −3);
• 3[2]−3[x−3]−3[y−4]−3y−2 — distribute the exponent −3 to the factors 2, x−3, and y−4;
• 3[1/8]x9y12y−2 for x ≠ 0 — work out 2−3 and multiply exponents to simplify powers of powers;
• (3/8)x9y10 for x, y ≠ 0 — combine the constant factors and the y factors.
So in summary, 3[2(xy)−4x]−3y−2 ≡ (3/8)x9y10 for xy ≠ 0. Notice that I had to add in extra conditions on x and y whenever a negative exponent disappeared. If either x or y is zero, then the original expression is undefined, but the simplified expression would be defined (as zero) except that I included those conditions. This sort of complication never arises in actual monomials, where all of the exponents are whole numbers.
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