Sometimes, we want to multiply a number by itself several times. For example, if we are trying to guess someone’s password, we may want to know how many 5-letter sequences are possible. Following our lessons on combinatorics (see here), we know that there are 26 \times 26 \times 26 \times 26 \times 26 possibilities. But we may want an easier way to write this down. The solution is to use exponents. In general,

Rule for Positive Exponents

Using this rule, we can rewrite our earlier expression as simply 26^5. We call 26 the base and we call 5 the exponent or power. And in general, for b^n, b is the base and n is the exponent. In English, we say this is “b to the n-th power.” So we would say that 26^5 is 26 to the 5th power.

Now, we can take any number and raise it to the power of any positive integer simply by multiplying the number by itself that many times. So, we have:

We can even do this for irrational numbers, like \pi:

But we can also have negative exponents. When we take a negative exponent, we use the following formula:

Rule for Negative Exponents

    \[b^{-n} = (\frac{1}{b})^n\]

In essence, we flip the base and then take a normal exponent. Here are some examples:

    \[3^{-4} = (\frac{1}{3})^4 = \frac{1}{81}\]

    \[\frac{1}{8}^{-3} = \frac{8}{1}^3 = 8^3 = 192\]

    \[\pi^{-99} = \Big(\frac{1}{\pi}\Big)^{99}\]

Finally, if we raise anything (other than 0) to the 0th power, we get 1. In other words,

Rule for the 0th Power

If b \neq 0, then b^0 = 1.

For example:

    \[29^0 = 1\]

    \[\pi^0 = 1\]

    \[(49+28/3 + \sqrt{2} + \sqrt{3} + \sqrt{4892})^0 = 1\]

Now, there are some rules for how different things with exponents can be combined.

Multiplication and Division Rules for Exponents

    \[a^{b}  a^c = a^{b + c}\]

    \[\frac{a^b}{a^c} = a^{b - c}\]

Exponent Rule for Exponents

    \[(a^b)^c = a^{bc}\]

Practice Problems

Simplify the following expressions:

  1. 5^{20} \times 5^{27}
  2. \frac{5^3}{5^{15}}

  3. \frac{9^3}{3^2}
  4. (4^2)^3
  5. (28^3)^0

  6. (7^{-3})^{-4} + \frac{5^3}{25}

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