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Negative Exponents

Any value to the zero power equals 1, convert negative exponents

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Zero and Negative Exponents

How can you use the quotient rules for exponents to understand the meaning of a zero or negative exponent?

Zero and Negative Exponents

Zero Exponent

Recall that \begin{align*}\boxed{\frac{a^m}{a^n}=a^{m-n}}\end{align*}. If \begin{align*}m = n\end{align*}, then the following would be true:

\begin{align*}\frac{a^m}{a^n}&=a^{{\color{red}m-n}}=a^{\color{red}0}\\ \frac{3^3}{3^3} &= 3^{{\color{red}3-3}}=3^{\color{red}0}\end{align*}

However, any quantity divided by itself is equal to one. Therefore, \begin{align*}\frac{3^3}{3^3}=1\end{align*} which means \begin{align*}3^{\color{red}0}={\color{red}1}\end{align*}. This is true in general:

\begin{align*}\boxed{a^{\color{red}0}=1 \ \text{if} \ a \neq 0.}\end{align*}

Note that if \begin{align*}a=0, \ 0^{\color{red}0}\end{align*} is not defined.

Negative Exponents

\begin{align*}4^2 \times 4^{-2}=4^{\color{red}2+(-2)}=4^{\color{red}0}={\color{red}1}\end{align*}

Therefore:

\begin{align*}& 4^2 \times 4^{-2}=1\\ & \frac{4^2 \times 4^{-2}}{4^2}=\frac{1}{4^2} && \text{Divide both sides by} \ 4^2.\\ & \frac{\cancel{4^2} \times 4^{-2}}{\cancel{4^2}}=\frac{1}{4^2} && \text{Simplify the equation.}\\ & \boxed{4^{{\color{red}-2}}=\frac{1}{4^{\color{red}2}}}\end{align*}

This is true in general and creates the following laws for negative exponents:

  • \begin{align*}\boxed{a^{{\color{red}-m}}=\frac{1}{a^{\color{red}m}}}\end{align*}
  • \begin{align*}\boxed{\frac{1}{a^{{\color{red}-m}}}=a^{\color{red}m}}\end{align*}

These laws for negative exponents can be expressed in many ways:

  • If a term has a negative exponent, write it as 1 over the term with a positive exponent. For example: \begin{align*}a^{\color{red}-m}=\frac{1}{a^{\color{red}m}}\end{align*} and \begin{align*}\frac{1}{a^{\color{red}-m}}=a^{\color{red}m}\end{align*}
  • If a term has a negative exponent, write the reciprocal with a positive exponent. For example: \begin{align*}\left(\frac{2}{3}\right)^{{\color{red}-2}}=\left(\frac{3}{2}\right)^{\color{red}2}\end{align*} and \begin{align*}a^{{\color{red}-m}}=\frac{a^{-m}}{1}=\frac{1}{a^{\color{red}m}}\end{align*}
  • If the term is a factor in the numerator with a negative exponent, write it in the denominator with a positive exponent. For example: \begin{align*}3x^{{\color{red}-3}}y=\frac{3y}{x^{\color{red}3}}\end{align*} and \begin{align*}a^{{\color{red}-m}}b^n=\frac{1}{a^{\color{red}m}}(b^n)=\frac{b^n}{a^{\color{red}m}}\end{align*}
  • If the term is a factor in the denominator with a negative exponent, write it in the numerator with a positive exponent. For example: \begin{align*}\frac{2x^3}{x^{-2}}=2x^3(x^2)\end{align*} and \begin{align*}\frac{b^n}{a^{{\color{red}-m}}}=b^n \left(\frac{a^{{\color{red}m}}}{1}\right)=b^na^{\color{red}m}\end{align*}

These ways for understanding negative exponents provide shortcuts for arriving at solutions without doing tedious calculations. The results will be the same.

Evaluate the exponents

Evaluate the following using the laws of exponents.

\begin{align*}\left(\frac{3}{4}\right)^{-2}\end{align*}

There are two methods that can be used to evaluate the expression.

Method 1: Apply the negative exponent rule \begin{align*}\boxed{a^{-m}=\frac{1}{a^m}}\end{align*}

\begin{align*}& \left(\frac{3}{4}\right)^{-2}=\frac{1}{{\color{red}\left(\frac{3}{4}\right)^2}} && \text{Write the expression with a positive exponent by applying} \ \boxed{a^{-m}=\frac{1}{a^m}}.\\ & \frac{1}{\left(\frac{3}{4}\right)^2}=\frac{1}{{\color{red}\frac{3^2}{4^2}}} && \text{Apply the law of exponents for raising a quotient to a power.} \ \boxed{\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}}\\ & \frac{1}{\frac{3^2}{4^2}}=\frac{1}{{\color{red}\frac{9}{16}}} && \text{Evaluate the powers.}\\ & \frac{1}{\frac{9}{16}}=1 \div \frac{9}{16} && \text{Divide}\\ & 1 \div \frac{9}{16}=1 \times \frac{16}{9}={\color{red}\frac{16}{9}}\\ & \boxed{\left(\frac{3}{4}\right)^{-2}=\frac{16}{9}}\end{align*}

Method 2: Apply the shortcut and write the reciprocal with a positive exponent.

\begin{align*}& \left(\frac{3}{4}\right)^{-2}={\color{red}\left(\frac{4}{3}\right)^2} && \text{Write the reciprocal with a positive exponent.}\\ & \left(\frac{4}{3}\right)^2={\color{red}\frac{4^2}{3^2}} && \text{Apply the law of exponents for raising a quotient to a power.} && \boxed{\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}}\\ & \frac{4^2}{3^2}={\color{red}\frac{16}{9}} && \text{Simplify.}\\ & \boxed{\left(\frac{3}{4}\right)^{-2}=\frac{16}{9}}\end{align*}

Applying the shortcut facilitates the process for obtaining the solution.

State the following using only positive exponents: (If possible, use shortcuts)

i) \begin{align*}y^{-6}\end{align*}

\begin{align*}& y^{-6} && \text{Write the expression with a positive exponent by applying} && \boxed{a^{-m}=\frac{1}{a^m}}.\\ & \boxed{y^{-6}=\frac{1}{y^6}}\end{align*}

ii) \begin{align*}\left(\frac{a}{b}\right)^{-3}\end{align*}

\begin{align*}& \left(\frac{a}{b}\right)^{-3} && \text{Write the reciprocal with a positive exponent.}\\ & \left(\frac{a}{b}\right)^{-3}={\color{red}\left(\frac{b}{a}\right)^3} && \text{Apply the law of exponents for raising a quotient to a power.} && \boxed{\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}}\\ & \left(\frac{b}{a}\right)^3={\color{red}\frac{b^3}{a^3}}\\ & \boxed{\left(\frac{a}{b}\right)^{-3}=\frac{b^3}{a^3}}\end{align*}

iii) \begin{align*}\frac{x^5}{y^{-4}}\end{align*}

\begin{align*}& \frac{x^5}{y^{-4}} && \text{Apply the negative exponent rule.} \ \boxed{\frac{1}{a^{{\color{red}-m}}}=a^{\color{red}m}}\\ & \frac{x^5}{y^{-4}}=x^5 \left(\frac{y^{\color{red}4}}{1}\right) && \text{Simplify}.\\ & \boxed{\frac{x^5}{y^{-4}}=x^5 y^4}\end{align*}

iv) \begin{align*}a^2 \times a^{-5}\end{align*}

\begin{align*}& a^2 \times a^{-5} && \text{Apply the product rule for exponents} \ \boxed{a^m \times a^n=a^{m+n}}.\\ & a^2 \times a^{-5}=a^{{\color{red}2+(-5)}} && \text{Simplify}.\\ & a^{2+(-5)}=a^{{\color{red}-3}} && \text{Write the expression with a positive exponent by applying} \ \boxed{a^{-m}=\frac{1}{a^m}}.\\ & a^{-3}={\color{red}\frac{1}{a^3}}\\ & \boxed{a^2 \times a^{-5}=\frac{1}{a^3}}\end{align*}

Evaluate

Evaluate the following: \begin{align*}\frac{7^{-2}+7^{-1}}{7^{-3}+7^{-4}}\end{align*}

There are two methods that can be used to evaluate the problem.

Method 1: Work with the terms in the problem in exponential form.

Numerator:

\begin{align*}& 7^{-2}=\frac{1}{7^2} \ \text{and} \ 7^{-1}=\frac{1}{7} && \text{Apply the definition} \ a^{-m}=\frac{1}{a^m}\\ & \frac{1}{7^2}+\frac{1}{7} && \text{A common denominator is needed to add the fractions.}\\ & \frac{1}{7^2}+\frac{1}{7} {\color{red}\left(\frac{7}{7}\right)} && \text{Multiply} \ \frac{1}{7} \ \text{by} \ \frac{7}{7} \ \text{to obtain the common denominator of} \ 7^2\\ & \frac{1}{7^2}+\frac{{\color{red}7}}{7^2}=\frac{1+7}{7^2}={\color{red}\frac{8}{7^2}} && \text{Add the fractions.}\end{align*}

Denominator:

\begin{align*}& 7^{-3}=\frac{1}{7^3} \ \text{and} \ 7^{-4}=\frac{1}{7^4} && \text{Apply the definition} \ a^{-m}=\frac{1}{a^m}\\ & \frac{1}{7^3}+\frac{1}{7^4} && \text{A common denominator is needed to add the fractions.}\\ & {\color{red}\left(\frac{7}{7}\right)} \frac{1}{7^3}+\frac{1}{7^4} && \text{Multiply} \ \frac{1}{7^3} \ \text{by} \ \frac{7}{7} \ \text{to obtain the common denominator of} \ 7^4\\ & \frac{{\color{red}7}}{7^4}+\frac{1}{7^4}=\frac{1+{\color{red}7}}{7^4}={\color{red}\frac{8}{7^4}} && \text{Add the fractions.}\end{align*}

Numerator and Denominator:

\begin{align*}& \frac{8}{7^2} \div \frac{8}{7^4} && \text{Divide the numerator by the denominator.}\\ & \frac{8}{7^2} \times \frac{7^4}{8} && \text{Multiply by the reciprocal.}\\ & \frac{\cancel{8}}{7^2} \times \frac{7^4}{\cancel{8}}=\frac{7^4}{7^2}=7^{\color{red}2}={\color{red}49} && \text{Simplify.}\\ & \boxed{\frac{7^{-2}+7^{-1}}{7^{-3}+7^{-4}}=49}\end{align*}

Method 2: Multiply the numerator and the denominator by \begin{align*}7^4\end{align*}. This will change all negative exponents to positive exponents. Apply the product rule for exponents and work with the terms in exponential form.

\begin{align*}& \frac{7^{-2}+7^{-1}}{7^{-3}+7^{-4}}\\ & {\color{red}\left(\frac{7^4}{7^4}\right)} \frac{7^{-2}+7^{-1}}{7^{-3}+7^{-4}} && \text{Apply the distributive property with the product rule for exponents.}\\ & \frac{7^{\color{red}2}+7^{\color{red}3}}{7^{\color{red}1}+7^{\color{red}0}} && \text{Evaluate the numerator and the denominator.}\\ & \frac{49+343}{7+1}=\frac{392}{8}={\color{red}49}\\ & \boxed{\frac{7^{-2}+7^{-1}}{7^{-3}+7^{-4}}=49}\end{align*}

Whichever method is used, the result is the same.

Examples

Example 1

Earlier, you were asked how can you use the quotient rules for exponents to understand the meaning of a zero or negative exponent?

By the quotient rule for exponents, \begin{align*}\frac{x^m}{x^m}=x^{m-m}=x^0\end{align*}. Since anything divided by itself is equal to 1 (besides 0), \begin{align*}\frac{x^m}{x^m}=1\end{align*}. Therefore, \begin{align*}x^0=1\end{align*} as long as \begin{align*}x\neq 0\end{align*}.

Also by the quotient rule for exponents, \begin{align*}\frac{x^2}{x^5}=x^{2-5}=x^{-3}\end{align*}. If you were to expand and reduce the original expression you would have \begin{align*}\frac{x^2}{x^5}=\frac{x\cdot x}{x\cdot x \cdot x\cdot x \cdot x}=\frac{1}{x^3}\end{align*}. Therefore, \begin{align*}x^{-3}=\frac{1}{x^3}\end{align*}. This generalizes to \begin{align*}x^{-a}=\frac{1}{x^a}\end{align*}.

Example 2

Use the laws of exponents to simplify the following: \begin{align*}(-3x^2)^3 (9x^4y)^{-2}\end{align*}

\begin{align*}& (-3x^2)^3(9x^4y)^{-2} && \text{Apply the laws of exponents} \ \boxed{(a^m)^n=a^{mn}} \ \text{and} \ \boxed{a^{-m}=\frac{1}{a^m}}\\ & (-3x^2)^3 (9x^4y)^{-2}=(-3^{\color{red}3}x^{\color{red}6}) \left(\frac{1}{(9x^4y)^2}\right) && \text{Simplify and apply} \ \boxed{(ab)^n=a^nb^n}\\ & (-3^3x^6) \left(\frac{1}{(9x^4y)^2}\right)={\color{red}-27}x^6 \left(\frac{1}{(9^{\color{red}2} x^{\color{red}8} y^{\color{red}2})}\right) && \text{Simplify}.\\ & -27x^6 \left(\frac{1}{(9^2x^8y^2)}\right)=\frac{-27x^6}{{\color{red}81}x^8y^2} && \text{Simplify and apply the quotient rule for exponents } \boxed{\frac{a^m}{a^n}=a^{m-n}}.\\ & \frac{-27x^6}{81x^8y^2}={\color{red}-\frac{1x^{-2}}{3y^2}} && \text{Apply the negative exponent rule} \ \boxed{a^{-m}=\frac{1}{a^m}}\\ & \boxed{(-3x^2)^3 (9x^4y)^{-2}=-\frac{1}{3x^2y^2}}\end{align*}

Example 3

Rewrite the following using only positive exponents. \begin{align*}(x^2 y^{-1})^2\end{align*}

\begin{align*} (x^2 y^{-1})^2 &= x^4y^{-2}\\ &=\frac{x^4}{y^2}\end{align*}

Example 4

Use the laws of exponents to evaluate the following: \begin{align*}[5^{-4} \times (25)^3]^2\end{align*}

\begin{align*}& [5^{-4} \times (25)^3]^2 && \text{Try to do this one by applying the laws of exponents.}\\ & [5^{-4} \times (25)^3]^2=[5^{-4} \times ({\color{red}5^2})^3]^2\\ & [5^{-4} \times ({\color{red}5^2})^3]^2=[5^{-4} \times 5^{\color{red}6}]^2\\ & [5^{-4} \times 5^{\color{red}6}]^2=(5^{\color{red}2})^2\\ & (5^{\color{red}2})^2=5^{\color{red}4}\\ & 5^4={\color{red}625}\\ & \boxed{[5^{-4} \times (25)^3]^2=5^4=625}\end{align*}

Review

Evaluate each of the following expressions:

  1. \begin{align*}-\left(\frac{2}{3}\right)^0\end{align*}
  2. \begin{align*}\left(-\frac{2}{5}\right)^{-2}\end{align*}
  3. \begin{align*}(-3)^{-3}\end{align*}
  4. \begin{align*}6 \times \left(\frac{1}{2}\right)^{-2}\end{align*}
  5. \begin{align*}7^{-4} \times 7^4\end{align*}

Rewrite the following using positive exponents only. Simplify where possible.

  1. \begin{align*}(4wx^{-2}y^3z^{-4})^3\end{align*}
  2. \begin{align*}\frac{a^2b^3c^{-2}}{d^{-2}bc^{-6}}\end{align*}
  3. \begin{align*}x^{-2}(x^2-1)\end{align*}
  4. \begin{align*}m^4(m^2+m-5m^{-2})\end{align*}
  5. \begin{align*}\frac{x^{-2}y^{-2}}{x^{-1}y^{-1}}\end{align*}
  6. \begin{align*}\left(\frac{x^{-2}}{y^4}\right)^3\left(\frac{y^{-4}}{x^6}\right)^{-7}\end{align*}
  7. \begin{align*}\frac{(x^{-2}y^4)^2}{(x^5y^{-3})^4}\end{align*}
  8. \begin{align*}\frac{(3xy^2)^3}{(3x^2y)^4}\end{align*}
  9. \begin{align*}\left(\frac{x^2y^{-25}z^5}{-12.4x^3y}\right)^0\end{align*}
  10. \begin{align*}\left(\frac{x^{-2}}{y^3}\right)^5\left(\frac{y^{-2}}{x^4}\right)^{-3}\end{align*}

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 6.4. 

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Vocabulary

Negative Exponent Property

The negative exponent property states that \frac{1}{a^m} = a^{-m} and \frac{1}{a^{-m}} = a^m for a \neq 0.

quotient rule

In calculus, the quotient rule states that if f and g are differentiable functions at x and g(x) \ne 0, then \frac {d}{dx}\left [ \frac{f(x)}{g(x)} \right ]= \frac {g(x) \frac {d}{dx}\left [{f(x)} \right ] - f(x) \frac{d}{dx} \left [{g(x)} \right ]}{\left [{g(x)} \right ]^2}.

Zero Exponent Property

The zero exponent property says that for all a \neq 0, a^0 = 1.

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