The exponent rule for negative exponents says that any power with a negative exponent is equal to the reciprocal of the power with a positive exponent.

\(x^{-a}=\frac{1}{x^{a}}\)

\(2^{-3}=\frac{1}{2^{3}}\)

\((-4)^{-5}=\frac{1}{(-4)^{5}}\)

\(x^{-7}\frac{1}{x^{7}}\)

- Write the reciprocal of the power with a positive exponent.
- Simplify the positive exponent, if possible.

Simplify \(5^{-6}2^{3}x^{2}y^{-4}\).

Each of the negative exponents can be simplified by writing the reciprocal with a positive exponent.

The powers with positive exponents will stay in the numerator.

\(5^{-6}2^{3}x^{2}y^{-4}=\frac{2^{3}x^{2}}{5^{6}y^{4}}\)

The \(2^3\) and \(5^6\) can be simplified.

\(2^{3}=2\times2\times2=8\)

\(5^{6}=5\times5\times5\times5\times5\times5=15,625\)

The \(x^2\) and \(y^4\) are as simplified as they can be, so...

\(\frac{2^{3}x^{2}}{5^{6}y^{4}}=\frac{8x^{2}}{15625y^{4}}\)

Answer: \(5^{-6}2^{3}x^{2}y^{-4} = \frac{8x^{2}}{15625y^{4}}\)

If you have an expression with multiple exponents or multiple bases, you can use any of the exponent rules to simplify the negative exponents.

Powers with an exponent of 0 always simplify to equal 1.

If you extend that reasoning, you'll see why negative exponents can be simplified as reciprocals with positive exponents.

To see how it works, let's write out the powers of 5 and the powers of 2 and look for patterns...

\(5^4=5\times5\times5\times5=625\)

\(5^3=5\times5\times5=125\)

\(5^2=5\times5=25\)

\(5^1=5\)

\(5^0=1\)

\(5^{-1}=\) **?**

\(5^{-2}=\) **?**

\(5^{-3}=\) **?**

Notice the pattern...

\(625\div5=125\)

\(125\div5=25\)

\(25\div5=5\)

\(5\div5=1\)

\(1\div5=\frac{1}{5}\)

\(\frac{1}{5}\div5=\frac{1}{25}\)

\(\frac{1}{25}\div5=\frac{1}{125}\)

This pattern shows us that \(5^{-1}=\frac{1}{5}\). You could also say that it equals \(\frac{1}{5^1}\) because the 5 has an invisible exponent of 1.

Similarly, \(5^{-2}=\frac{1}{25}\), which could be written as \(\frac{1}{5^2}\).

And \(5^{-3}=\frac{1}{125}\), which could be written as \(\frac{1}{5^3}\).

\(2^4=2\times2\times2\times2=16\)

\(2^3=2\times2\times2=8\)

\(2^2=2\times2=4\)

\(2^1=2\)

\(2^0=1\)

\(2^{-1}=\) **?**

\(2^{-2}=\) **?**

\(2^{-3}=\) **?**

Notice the pattern...

\(16\div2=8\)

\(8\div2=4\)

\(4\div2=2\)

\(2\div2=1\)

\(1\div2=\frac{1}{2}\)

\(\frac{1}{2}\div2=\frac{1}{4}\)

\(\frac{1}{4}\div2=\frac{1}{8}\)

This pattern tells us that...

\(2^{-1}=\frac{1}{2^1}\)

\(2^{-2}=\frac{1}{2^2}\)

\(2^{-3}=\frac{1}{2^3}\)

A similar pattern will show up if you write out the powers of any other number. So, that's why negative exponents are equal to the reciprocal power with a positive exponent.

If you have negative exponents in the denominator of a fraction, the reciprocal will move the power up to the numerator.

\(\frac{1}{2^{-5}}=2^{5}\)

\(\frac{1}{a^{-8}}=a^{8}\)

\(\frac{x^{-4}}{y^{-3}}=\frac{y^3}{x^4}\)

The power moves to the numerator because of the way that fraction division works.

Simplify \(\frac{1}{7^{-3}}\)

The fraction bar is a division bar, so we can re-write the problem as...

\(\frac{1}{7^{-3}}=1\div7^{-3}\)

When we simplify the negative exponent, we get...

\(\frac{1}{7^{-3}}=1\div\frac{1}{7^{3}}\)

Next, we'll multiply by the reciprocal to finish the fraction division.

\(1\div\frac{1}{7^{3}}=1\times\frac{7^3}{1}=7^{3}\)

Answer: \(\frac{1}{7^{-3}}=7^{3}\)

You can only move the power from the denominator to the numerator if ** everything** in the denominator is

If there's any addition or subtraction in the denominator, you have to write the reciprocal like this...

\(\frac{1}{5+3^{-2}}=\frac{1}{5+\frac{1}{3^{2}}}\)

In this example, the -2 exponent only applies to the 3 so the reciprocal would be \(\frac{1}{3^2}\). However, that reciprocal has to be added to the 5 in the denominator, so it does not move up to the numerator.

Khan Academy - Negative Exponents

Homeschool Math - Whole Number Bases

Homeschool Math - Fraction and Decimal Bases

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