Negative Integer Exponents

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Explore This 2 Summary

In this exploration, you had a chance to look at the effect that a negative exponent has on a positive base. You have probably noticed that when an exponent is negative, the power is small and can be expressed as a fraction.

The value of both examples is a fraction! Let's look at the math behind it.

Example 3

Solution

These solutions result in different looking answers. We know that our math is correct using both methods, so it must mean that these two final answers are in fact equivalent, \(2^{-3}=\dfrac{1}{2^3}\).

Convince yourself that the result would be the same with a different base (except \(0\)) and with different integer exponents. 

It is important to notice two things from the rule:

• the base stays the same, and
• the exponent becomes positive when the power is moved from the numerator to the denominator.

What happens when the negative exponent is in the denominator to start?

The base stayed the same, and the exponent became positive when the power moved from the denominator to the numerator.

Example 4

Solution:

1. For \(2^{-1}\), the exponent \(-1\) has a base of \(2\). The power with a negative exponent is in the numerator. To write the power with a positive exponent, keep the base the same, change the exponent to positive and move to the denominator.\[\begin{align*} 2^{-1}&=\frac{1}{2^1}\\ &=\frac{1}{2} \end{align*}\]

   Notice that \(2\) and \(\dfrac{1}{2}\) are reciprocals of one another. Since \(2^{-1}=\dfrac{1}{2}\), an exponent of \(-1\) produces the reciprocal of the base, in this case \(2\).

2. For \(\left(\dfrac{2}{3}\right)^{-1}\), the exponent of \(-1\) applies to everything inside of the bracket. To write this with a positive exponent, keep the base the same, change the exponent to positive, and move \(3^1\) to the numerator and \(2^1\) to the denominator.\[\begin{align*} \left(\frac{2}{3}\right)^{-1}&=\frac{2^{-1}}{3^{-1}}\\ &=\frac{3^1}{2^1}\\ &=\frac{3}{2} \end{align*}\]Again, notice that \(\dfrac{2}{3}\) and \(\dfrac{3}{2}\) are reciprocals of one another, so an exponent of \(-1\) produces the reciprocal of the base.
3. For \(8^{-2}\), the exponent \(-2\) has a base of \(8\). The power with a negative exponent is in the numerator. To write the power with a positive exponent, keep the base the same, change the exponent to positive, and move to the denominator.\[8^{-2}=\frac{1}{8^2}\]The question stated to write with positive exponents, so we stop here rather than evaluating \(8^2\).
4. For \((-3)^{-5}\), the exponent \(-5\) has a base of \((-3)\). The power with a negative exponent is in the numerator. To write the power with a positive exponent, keep the base the same, change the exponent to positive, and move to the denominator.\[(-3)^{-5}=\frac{1}{(-3)^5}\]The question stated to write with positive exponents, so we stop here rather than evaluating \((-3)^5\).
5. For \(\dfrac{1}{x^{-4}}\), we will have two approaches for writing with positive exponents:
    

   Approach 1: Division of fractions

   Using division of fractions we get that

   \[\begin{align*} \frac{1}{x^{-4}}&=\frac{1}{\frac{1}{x^4}}\\ &=1\div\frac{1}{x^4}\\ &=1\times x^4\\ &=x^4 \end{align*}\]

   Approach 2: Exponent Law

   The power with a negative exponent is in the denominator. To write the power with a positive exponent, keep the base the same, change the exponent to positive, and move to the numerator, so \(\dfrac{1}{x^{-4}}=x^4\).

Example 5

Solution

1. For \(5x^{-7}\), the exponent \(-7\) has a base of \(x\), while \(5\) has an exponent of \(1\). Since \(5\) has a positive exponent but \(x\) does not, then only \(x^7\)moves to the denominator. Therefore,\[\begin{align*} 5x^{-7}&=5^1x^{-7}\\ &=5\left(\frac{1}{x^7}\right)\\ &=\frac{5}{x^7} \end{align*}\]
2. For \((2m^{-8})(5m^3)\), when asked to simplify, follow the order of BEDMAS. Using the exponent rules previously learned and then ensuring that our answer has positive exponents, we get\[\begin{align*} (2m^{-8})(5m^3)&=10m^{-8+3}\\ &=10m^{-5}\\ &=10\left(\frac{1}{m^5}\right)\\ &=\frac{10}{m^5} \end{align*}\]
3. For \(\dfrac{49a^5b^2}{7a^2b^6}\), using BEDMAS and previously learned exponent rules, we get that\[\begin{align*} \frac{49a^5b^2}{7a^2b^6}&=\frac{49}{7}a^{5-2}b^{2-6}\\ &=7a^3b^{-4}\\ &=7a^3\left(\frac{1}{b^4}\right)\\ &=\frac{7a^3}{b^4} \end{align*}\]
4. For \(\dfrac{(2x^2y^4)^2}{2^3x^6y^3}\), using BEDMAS and previously learned exponent rules, we get that\[\begin{align*} \frac{(2x^2y^4)^2}{2^3x^6y^3}&=\frac{2^2(x^2)^2(y^4)^2}{2^3x^6y^3}\\ &=\frac{2^2x^4y^8}{2^3x^6y^3}\\ &=2^{2-3}x^{4-6}y^{8-3}\\ &=2^{-1}x^{-2}y^5\\ &=\frac{1}{2}\frac{1}{x^2}y^5\\ &=\frac{y^5}{2x^2} \end{align*}\]

Try This Revisited

Solution

1. The length of ten red blood cells can be found by multiplying the length of one blood cell by \(10\) and then simplifying:\[\begin{align*} 8.5\times 10^{-3}\times 10&=8.5\times 10^{-3+1}\\ &=8.5\times 10^{-2} \end{align*}\]Therefore, the length of ten red blood cells is \(8.5\times 10^{-2}\) millimetres.
2. What about one hundred red blood cells (in millimetres)?

   Start by thinking of \(100\) as \(10^2\).

   \[\begin{align*} 8.5\times 10^{-3}\times 10^2&=8.5\times 10^{-3+2}\\ &=8.5\times 10^{-1} \end{align*}\]

   Therefore, the length of \(100\) red blood cells is \(8.5\times 10^{-1}\) millimetres.