We have seen how we can rewrite expressions with negative powers to have a positive powers
For example, if we simplified $a^3\div a^5$a3÷a5 using the division law, we would get $a^{-2}$a−2. Let's expand the example to see why this is the case:
Remember that when we are simplifying fractions, we are looking to cancel out common factors in the numerator and denominator. Remember that any number divided by itself is $1$1.
So using the second approach, we can also express $a^3\div a^5$a3÷a5 with a positive exponent as $\frac{1}{a^2}$1a2. This gives us the negative exponent law. When dealing with algebraic bases we follow exact the same approach.
For any base $a$a,
$a^{-x}=\frac{1}{a^x}$a−x=1ax, $x\ne0$x≠0.
That is, when raising a base to a negative power:
When raising a fractional base to a negative power we can combine the individual rules we have seen.
Express the following with a positive exponent: $\left(\frac{a}{b}\right)^{-3}$(ab)−3
Think: We want to combine the rules for raising fractions with the rule for negative exponents.
That is we want to use the rules $\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$(ab)n=anbn and $a^{-n}=\frac{1}{a^n}$a−n=1an.
Do:
$\left(\frac{a}{b}\right)^{-3}$(ab)−3 | $=$= | $\frac{a^{-3}}{b^{-3}}$a−3b−3 |
Use the rule for raising a fraction |
$=$= | $a^{-3}\div b^{-3}$a−3÷b−3 |
Rewrite the quotient with a division symbol |
|
$=$= | $\frac{1}{a^3}\div\frac{1}{b^3}$1a3÷1b3 |
Apply the negative exponent rule to the numerator and the denominator to express both with positive exponents |
|
$=$= | $\frac{1}{a^3}\times\frac{b^3}{1}$1a3×b31 |
Dividing by a fraction is the same as multiplying by the reciprocal of that fraction |
|
$=$= | $\frac{b^3}{a^3}$b3a3 |
Simplify the fractional product |
|
$=$= | $\left(\frac{b}{a}\right)^3$(ba)3 |
Write as a single term raised to a power by using the reverse of the rule for raising fractions |
We can see that $\left(\frac{a}{b}\right)^{-3}=\frac{b^3}{a^3}$(ab)−3=b3a3$=$=$\left(\frac{b}{a}\right)^3$(ba)3
Reflect: What has happened is we have found the reciprocal of the expression in the question, and turned the power into a positive. Using this trick will save a lot of time!
For any base number of the form $\frac{a}{b}$ab, and any number $n$n as a power,
$\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$(ab)n=anbn
If $n$n is negative, then we also use the fact $a^{-n}=\frac{1}{a}$a−n=1a. Giving us the following rule:
$\left(\frac{a}{b}\right)^{-n}=\left(\frac{b}{a}\right)^n$(ab)−n=(ba)n
Find the value of $n$n such that $\frac{1}{25}=5^n$125=5n.
Simplify the following, giving your answer with a positive exponent:
$\frac{9x^2}{3x^9}$9x23x9Simplify the following, giving your answer with a positive exponent:
$\left(\frac{y}{4}\right)^{-3}$(y4)−3