1.03 Negative exponents
The negative power property 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.
Worked example
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 power property, 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.
Raise 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.
Worked example
Question 2
Calculate $\left(\frac{a}{b}\right)^{-3}$(ab)−3
Think:
| $\left(\frac{a}{b}\right)^{-3}$(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. |
Solve: $\left(\frac{a}{b}\right)^{-3}=\frac{b^3}{a^3}$(ab)−3=b3a3
Reflect: Can you see 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.
Multiplication and division properties with negative exponents
We've already learned about the properties for exponents with multiplication and division, as well as the negative power property. Now we are going to combine these rules to simplify expressions which involve multiplication or division, and negative exponents.
Consider the expression: $e^7\times e^{-4}$e7×e−4
Notice the following:
- There is multiplication and the bases are the same (we can apply the multiplication law)
- One of the powers is negative (we can express the second term with a positive power if we wish)
When negative powers are involved, this opens up choices in how we go about trying to simplify the expression.
With the above example, I have two choices:
One Approach: Add the powers immediately as the bases are the same and we are multiplying
| $e^7\times e^{-4}$e7×e−4 | $=$= | $e^{7+\left(-4\right)}$e7+(−4) | |
| $=$= | $e^{7-4}$e7−4 | (recall that a plus and minus sign next to each other result in a minus) | |
| $=$= | $e^3$e3 |
Another Approach: First express the second term with a positive power
| $e^7\times e^{-4}$e7×e−4 | $=$= | $e^7\times\frac{1}{e^4}$e7×1e4 | |
| $=$= | $\frac{e^7}{e^4}$e7e4 | ||
| $=$= | $e^{7-4}$e7−4 | (subtract the powers using the division rule) | |
| $=$= | $e^3$e3 |
Of course, which way you go about these types of problems is completely up to you.
Worked example
Question 3
Simplify the following, giving your answer in exponential form: $\frac{35x^2y^2}{-5x^{14}y^{14}}$35x2y2−5x14y14.
Think: Let's separate the terms and apply the Exponent Division law.
Do:
| $35\div\left(-5\right)$35÷(−5) | $=$= | $-7$−7 | the coefficient term |
| $x^2\div x^{14}$x2÷x14 | $=$= | $x^{2-14}$x2−14 | the $x$x's |
| $=$= | $x^{-12}$x−12 | the $x$x's | |
| $=$= | $\frac{1}{x^{12}}$1x12 | and take the reciprocal | |
| $y^2\div y^{14}$y2÷y14 | $=$= | $y^{2-14}$y2−14 | the $y$y's |
| $y^2\div y^{14}$y2÷y14 | $=$= | $y^{-12}$y−12 | the $y$y's |
| $=$= | $\frac{1}{y^{12}}$1y12 |
and take the reciprocal |
|
| $=$= | $-7\times\frac{1}{x^{12}}\times\frac{1}{y^{12}}$−7×1x12×1y12 | all together | |
| $\frac{35x^2y^2}{-5x^{14}y^{14}}$35x2y2−5x14y14 | $=$= | $\frac{-7}{x^{12}y^{12}}$−7x12y12 | and simplify |
Power of a power rule with negative exponents
We already learned to apply the power of a power rule to positive exponents. This rule states:
$\left(x^a\right)^b=x^{a\times b}$(xa)b=xa×b
Now we are going to explore what happens when we also include negative values in these kinds of questions.
If you think back to learning about multiplying and dividing by negative numbers, you'll remember that:
- Multiplying a positive and a negative value will give you a negative answer.
- Multiplying two negative numbers together gives you a positive answer.
Let's think for a minute about what happens if we multiply a negative value by itself more than twice.
Say we had the question $\left(-2\right)^3$(−2)3. This means $-2\times\left(-2\right)\times\left(-2\right)$−2×(−2)×(−2). If we simplify this, $-2\times\left(-2\right)=4$−2×(−2)=4 and $4\times\left(-2\right)=-8$4×(−2)=−8.
What about if we multiplied it by itself again to get the answer to $\left(-2\right)^4$(−2)4? We know that $\left(-2\right)^3=-8$(−2)3=−8 and $-8\times\left(-2\right)=16$−8×(−2)=16.
So, as a general rule:
- If you multiply a negative number by a positive power, you will get a positive answer
- If you multiply a negative number by a negative power, you will get a negative answer.
Worked example
Question 4
Simplify $\left(w^{10}\right)^{-5}$(w10)−5.
Think: Using the power to a power rule, we need to multiply the exponents.
$\left(w^{10}\right)^{-5}=w^{10\times\left(-5\right)}$(w10)−5=w10×(−5)
Do: $\left(w^{10}\right)^{-5}=w^{-50}$(w10)−5=w−50
Simplify the negative exponent: $\left(w^{10}\right)^{-5}=\frac{1}{w^{50}}$(w10)−5=1w50
Reflect: How would this problem be different if both powers were negative?
Practice questions
Question 1
Simplify the following, giving your answer with a positive exponent: $\frac{9x^3}{3x^{-4}}$9x33x−4
Question 2
Simplify the following, writing without negative exponents.
$5p^2q^{-4}\cdot8p^{-2}q^6$5p2q−4·8p−2q6
Question 3
Express $4y^{-2}\cdot2y^{-4}$4y−2·2y−4 with a positive exponent.
Question 4
Express $\left(\frac{x^2}{y^4}\right)^{-1}$(x2y4)−1 without negative exponents.