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
Analyse, through the use of patterning, the relationships between the exponents of powers and the operations with powers, and use these relationships to simplify numeric and algebraic expressions.
Simplify algebraic expressions by applying properties of operations of numbers, using various representations and tools, in different contexts.