We have looked before at how to manipulate negative exponents.
The negative exponent law states:
$a^{-x}=\frac{1}{a^x}$a−x=1ax
So if you need to express a negative exponent as a positive exponent, or a positive exponent as a negative exponent, you need to convert it to a fraction using this rule.
We now need to consider what happens if we have a coefficient as well as an unknown raised to a negative power.
Examples
Question 1
Express $3x^{-3}$3x−3 with a positive exponent.
Think - the $x$x has been raised to a power of $-3$−3. We must evaluate this first, then multiply by $3$3.
Do - Apply the negative exponent law, then multiply by $3$3.
| $3x^{-3}$3x−3 | $=$= | $3\times x^{-3}$3×x−3 |
| $=$= | $3\times\frac{1}{x^3}$3×1x3 | |
| $=$= | $\frac{3}{x^3}$3x3 |
As the power is only on the $x$x, the value of the coefficient hasn't changed.
A Fraction to a Negative Power
Things become a bit more complicated when we need to raise a fraction to a negative power.
First, you may want to remind yourself what we mean by a reciprocal.
The reciprocal of $\frac{1}{8}$18 is $8$8. The reciprocal of $\frac{2}{3}$23 is $\frac{3}{2}$32. The reciprocal of $91$91 is $\frac{1}{91}$191.
Can you see what reciprocal means? Simply, if you have a fraction, you need to invert it (or flip it) to find the reciprocal. If you have an integer, you put that integer as the denominator, and $1$1 as the numerator.
This links to negative powers, as we saw above.
Example
Question 2
So what happens if you are asked to calculate $\left(\frac{a}{b}\right)^{-3}$(ab)−3?
| $\left(\frac{a}{b}\right)^{-3}$(ab)−3 | $=$= | $\frac{a^{1\times\left(-3\right)}}{b^{1\times\left(-3\right)}}$a1×(−3)b1×(−3) | First we need to multiply out the powers. |
| $=$= | $\frac{a^{-3}}{b^{-3}}$a−3b−3 | We know that this is the same as | |
| $=$= | $a^{-3}\div b^{-3}$a−3÷b−3 | Now let's rewrite this expression using only positive powers | |
| $=$= | $\frac{1}{a^3}\div\frac{1}{b^3}$1a3÷1b3 | Dividing by a fraction is the same as multiplying by the reciprocal of that fraction | |
| $=$= | $\frac{1}{a^3}\times b^3$1a3×b3 | Can you see what is going to happen next? By multiplying through we now get | |
| $=$= | $\frac{b^3}{a^3}$b3a3 | Let's take a moment to compare this answer to the question. |
$\left(\frac{a}{b}\right)^{-3}=\frac{b^3}{a^3}$(ab)−3=b3a3
Look for a shortcut here.
What has happened is we have found the reciprocal of the question, and raised each term to the power which is now a positive.
Question 3
Now let's try another.
Express $\left(\frac{x^8}{y^5}\right)^{-4}$(x8y5)−4 with a positive exponent.
| $=$= | $\frac{x^{8\times\left(-4\right)}}{y^{5\times\left(-4\right)}}$x8×(−4)y5×(−4) | |
| $=$= | $\frac{x^{-32}}{y^{-20}}$x−32y−20 | |
| $=$= | $\frac{y^{20}}{x^{32}}$y20x32 |
Now it's your turn to practice applying these rules.
Worked Examples
Question 4
Express $4y^{-2}\times2y^{-4}$4y−2×2y−4 with a positive exponent.
Question 5
Express $\left(\frac{x^2}{y^4}\right)^{-1}$(x2y4)−1 without negative exponents.