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Exponent laws with variables

Algebra · level 6, lesson 1 of 6

A question to start

Without a calculator, write \(4^5 \times 8^2\) as a single power of 2.

It looks like you need to work out \(1024 \times 64\) and then find out which power of 2 that is. But there is a much quicker way. \(4 = 2^2\) and \(8 = 2^3\), so the question is really about multiplying lots of 2's together, and all you have to do is count them. The answer is \(2^{16}\), worked out in two lines in worked example 6.

The rules that let you count instead of multiply are the exponent laws. You may have met them with numbers; in this lesson they work with letters, and you will see exactly why each one is true, so you never need to memorise them blindly.

The idea

Powers, bases and exponents

A power is a short way to write repeated multiplication:

\[x^5 = x \times x \times x \times x \times x.\]

The repeated number \(x\) is the base, and the small raised number 5 is the exponent (or index). It counts how many copies of the base are multiplied. For now the exponents are positive whole numbers; zero and negative exponents come later in the lesson.

Every law below comes from one idea: write the powers out in full, and count the factors.

Law 1: multiplying powers with the same base

\[x^3 \times x^4 = (x \times x \times x) \times (x \times x \times x \times x) = x^7.\]
xxx × xxxx x³: 3 factorsx⁴: 4 factors = x⁷3 + 4 = 7

There are 3 factors of \(x\) and then 4 more, so 7 altogether. In general:

\[x^a \times x^b = x^{a+b}.\]

Multiplying powers of the same base adds the exponents. The base must be the same: \(x^3 \times y^4\) cannot be combined, because the factors are of different kinds.

Law 2: dividing powers with the same base

\[\frac{x^6}{x^2} = \frac{x \times x \times x \times x \times x \times x}{x \times x} = x \times x \times x \times x = x^4.\]

Each \(x\) on the bottom cancels one \(x\) on the top (because \(\tfrac{x}{x} = 1\), as long as \(x \ne 0\)). Two factors cancel, leaving \(6 - 2 = 4\). In general:

\[\frac{x^a}{x^b} = x^{a-b}, \qquad x \ne 0.\]

Dividing powers of the same base subtracts the exponents.

Law 3: a power of a power

\[(x^3)^4 = x^3 \times x^3 \times x^3 \times x^3 = x^{3+3+3+3} = x^{12}.\]

Four groups of 3 factors each is \(4 \times 3 = 12\) factors. In general:

\[(x^a)^b = x^{ab}.\]

A power of a power multiplies the exponents. Compare carefully with Law 1: \(x^3 \times x^4 = x^7\) (add), but \((x^3)^4 = x^{12}\) (multiply).

Law 4: a power of a product

\[(xy)^3 = (xy)(xy)(xy) = x \times x \times x \times y \times y \times y = x^3y^3.\]

Multiplication can be done in any order, so we can collect all the \(x\)'s together and all the \(y\)'s together. In general:

\[(xy)^n = x^n y^n.\]

This includes numbers: \((3x)^2 = 3^2x^2 = 9x^2\). The 3 is inside the bracket, so it is squared too. The same idea works for division: \(\left(\dfrac{x}{y}\right)^n = \dfrac{x^n}{y^n}\).

A warning: there is no law for sums

There is no law for \((x + y)^n\). In particular, \((x + y)^2\) is not \(x^2 + y^2\). Test with \(x = 1\) and \(y = 2\): \((1 + 2)^2 = 9\), but \(1^2 + 2^2 = 5\). Law 4 works because multiplying can be rearranged; a sum inside a bracket cannot be split up like that. What \((x + y)^2\) really equals is the subject of the next lesson.

Coefficients and several letters

When a term has a number and letters, deal with each part separately: numbers with numbers, \(x\)'s with \(x\)'s, \(y\)'s with \(y\)'s.

\[3x^2y \times 5x^4y^3 = (3 \times 5) \times x^{2+4} \times y^{1+3} = 15x^6y^4.\]

Remember that \(y\) on its own means \(y^1\).

Zero as an exponent

What should \(x^0\) mean? "Zero copies of \(x\) multiplied" sounds meaningless. But Law 2 gives an answer. Divide any power by itself:

\[\frac{x^3}{x^3} = x^{3-3} = x^0, \qquad\text{but also}\qquad \frac{x^3}{x^3} = 1.\]

For the law to stay true, we must define

\[x^0 = 1, \qquad x \ne 0.\]

A second way to see it: the list \(x^3, x^2, x^1\) is made by dividing by \(x\) each time. One more step down gives \(x^1 \div x = 1\), which is the next item, \(x^0\).

Negative exponents

The same pattern continues below zero. Dividing by \(x\) again and again:

\[x^2,\quad x^1,\quad x^0 = 1,\quad x^{-1} = \frac{1}{x},\quad x^{-2} = \frac{1}{x^2},\quad \ldots\]

and Law 2 agrees: \(\dfrac{x^2}{x^5} = x^{2-5} = x^{-3}\), while cancelling gives \(\dfrac{1}{x^3}\). So we define

\[x^{-n} = \frac{1}{x^n}, \qquad x \ne 0.\]

A negative exponent means "one over", not "negative": \(2^{-3} = \tfrac18\), a small positive number. With these definitions, all four laws keep working for every whole-number exponent, positive, zero or negative. That is the reason for choosing them. (The numbers version, with more examples, is in zero and negative exponents.)

Simplified form

An expression is usually called simplified when each base appears once, there are no brackets left, and every exponent is positive. So \(\dfrac{3}{x^2}\) is preferred to \(3x^{-2}\), though both are correct.

Worked examples

1. Law 1. \(x^4 \times x^7 = x^{11}\), and \(a \times a^5 \times a^2 = a^{1+5+2} = a^8\).

2. The outline's example. Simplify \(\dfrac{(2x^3y)^2}{4x^2y}\).

  • Top, by Laws 4 and 3: \((2x^3y)^2 = 2^2 \times (x^3)^2 \times y^2 = 4x^6y^2\).
  • Divide by Law 2, part by part: \(\dfrac{4}{4} = 1\), \(\dfrac{x^6}{x^2} = x^4\), \(\dfrac{y^2}{y} = y\).
  • Answer: \(x^4y\).
  • Check with \(x = 1\), \(y = 2\): top \((2 \times 1 \times 2)^2 = 16\), bottom \(4 \times 1 \times 2 = 8\), so \(16 \div 8 = 2\); and \(x^4y = 1 \times 2 = 2\) ✓.

3. A negative base. Simplify \((-3a^2b)^3\).

  • Cube each part: \((-3)^3 = -27\), \((a^2)^3 = a^6\), \(b^3\).
  • Answer: \(-27a^6b^3\). An odd power of a negative is negative.

4. Numbers that do not cancel fully. Simplify \(\dfrac{12a^5b^2}{8a^7b}\).

  • Numbers: \(\tfrac{12}{8} = \tfrac{3}{2}\). \(a\)'s: \(a^{5-7} = a^{-2} = \tfrac{1}{a^2}\). \(b\)'s: \(b^{2-1} = b\).
  • Answer: \(\dfrac{3b}{2a^2}\).

5. Negative exponents. Simplify \((x^{-2}y^3)^{-2}\), with positive exponents.

  • Law 3 on each part: \(x^{(-2)(-2)} = x^4\) and \(y^{3 \times (-2)} = y^{-6}\).
  • Answer: \(x^4y^{-6} = \dfrac{x^4}{y^6}\).

6. The question at the start. \(4^5 \times 8^2 = (2^2)^5 \times (2^3)^2 = 2^{10} \times 2^6 = 2^{16}\). Check: \(1024 \times 64 = 65536 = 2^{16}\) ✓.

Try it

  1. Simplify: (a) \(y^3 \times y^9\); (b) \(\dfrac{m^{10}}{m^4}\); (c) \((p^3)^5\).
  2. Simplify \((4x^2y^3)^2\).
  3. Simplify \(\dfrac{15x^6y^2}{5x^2y^2}\).
  4. Simplify \((2a^3)^2 \times 3a^4\).
  5. Write with positive exponents: (a) \(x^{-4} \times x^2\); (b) \((3x^{-1})^2\).
  6. Evaluate \(5^0\), \(2^{-3}\) and \(\left(\tfrac12\right)^{-2}\).
  7. Find \(n\) if \(2^n = \dfrac{8^3 \times 4^2}{2}\).
  8. True or false? Test each with \(x = 1\) and \(y = 2\). (a) \((x + y)^2 = x^2 + y^2\); (b) \((xy)^2 = x^2y^2\); (c) \(x^2 \times x^3 = x^6\).
Answers
  1. (a) \(y^{12}\) (add). (b) \(m^6\) (subtract). (c) \(p^{15}\) (multiply).
  2. \(4^2 \times x^{2 \times 2} \times y^{3 \times 2} = 16x^4y^6\).
  3. \(\tfrac{15}{5} = 3\), \(x^{6-2} = x^4\), \(y^{2-2} = y^0 = 1\). Answer: \(3x^4\).
  4. \((2a^3)^2 = 4a^6\), then \(4a^6 \times 3a^4 = 12a^{10}\).
  5. (a) \(x^{-4+2} = x^{-2} = \dfrac{1}{x^2}\). (b) \(3^2x^{-2} = \dfrac{9}{x^2}\).
  6. \(5^0 = 1\). \(2^{-3} = \tfrac{1}{8}\). \(\left(\tfrac12\right)^{-2} = \dfrac{1}{(1/2)^2} = \dfrac{1}{1/4} = 4\).
  7. \(8^3 = 2^9\) and \(4^2 = 2^4\), so the right side is \(\dfrac{2^9 \times 2^4}{2^1} = 2^{9+4-1} = 2^{12}\). So \(n = 12\). Check: \(512 \times 16 \div 2 = 4096 = 2^{12}\) ✓.
  8. (a) False: \(9 \ne 5\). (b) True: \((2)^2 = 4\) and \(1 \times 4 = 4\), and Law 4 says it is always true. (c) False: \(1 \times 1 = 1\) agrees by luck, since every power of 1 is 1, so test \(x = 2\): \(4 \times 8 = 32 = 2^5\), not \(2^6 = 64\). The correct answer is \(x^5\). (A test that agrees proves nothing; one that disagrees proves it false.)

Common mistake

Multiplying the bases or the exponents in Law 1. \(x^3 \times x^4\) is \(x^7\), not \(x^{12}\). And \(2^3 \times 2^4\) is \(2^7\), not \(4^7\): the base stays the same. Writing the factors out once, \((2 \times 2 \times 2)(2 \times 2 \times 2 \times 2)\), shows seven 2's.

Forgetting to raise the number. \((3x)^2 = 9x^2\), not \(3x^2\). But \(3x^2\) means \(3 \times x^2\): only \(x\) is squared. The bracket decides what the exponent applies to.

Where it leads