Learning Outcomes
- Explain the meaning of a zero exponent
- Explain the meaning of a negative exponent
- Apply the power of product rule for exponents
- Apply the power of quotient rule for exponents
- Simplify exponential expressions
- Use properties of exponents to write equivalent exponential functions in standard form
Zero Exponents
The quotient rule for exponents can be used to determine the meaning of [latex]x^0[/latex]:
The quotient rule for exponents tells us that by subtracting the exponents:
[latex]\dfrac{x^n}{x^n}=x^0[/latex]
But since the numerator and denominator are identical, we can cancel the terms by division:
[latex]\dfrac{x^n}{x^n}=1[/latex] provided [latex]x\neq0[/latex] since we can’t divide by 0.
Therefore,
[latex]x^0=1[/latex], [latex]x\neq0[/latex]
exponent of zero
For all real numbers [latex]a\neq0[/latex],
[latex]a^0=1[/latex]
For example,
[latex]\begin{aligned}2022^0&=1\\(pq)^0&=1,\;p,\;q\neq 0 \\(2050xy)^{0}&=1,\;x,\;y\neq 0\end{aligned}[/latex]
[latex]\begin{aligned}\dfrac{5 a^m z^2}{a^mz}&=5\cdot\dfrac{a^m}{a^m}\cdot\dfrac{z^2}{z}&&\text{Separate into fractions}\\&=5 \cdot a^{m-m}\cdot z^{2-1}&&\text{Subtract the exponents}\\&=5\cdot a^0 \cdot z^1 &&\text{Simplify }a^0=1\text{ and }z^1=z\\&=5z\end{aligned}[/latex]
Example 1
Simplify each expression.
- [latex]\dfrac{{c}^{3}}{{c}^{3}}[/latex]
- [latex]\dfrac{-3{x}^{5}}{{x}^{5}}[/latex]
- [latex]\dfrac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}[/latex]
- [latex]\dfrac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}[/latex]
Solution
We can apply the zero exponent rule and other rules to simplify each expression:
1. [latex]\begin{align}\frac{c^{3}}{c^{3}}&=c^{3-3}&&\text{Apply the quotient rule: subtract exponents} \\ & =c^{0}&&\text{Apply the zero exponent rule} \\ & =1\end{align}[/latex] |
2. [latex]\begin{align} \frac{-3{x}^{5}}{{x}^{5}}& = -3\cdot \frac{{x}^{5}}{{x}^{5}} \\ & = -3\cdot {x}^{5 - 5}&&\text{Apply the quotient rule: subtract exponents} \\ & = -3\cdot {x}^{0}&&\text{Apply the zero exponent rule}\\ & = -3\cdot 1 \\ & = -3 \end{align}[/latex] |
3. [latex]\begin{align} \frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}& = \frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{1+3}} && \text{Use the product rule in the denominator. The base is }(j^2k). \\ & = \frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{4}} && \text{Use the quotient rule}. \\ & = {\left({j}^{2}k\right)}^{4 - 4} \\ & = {\left({j}^{2}k\right)}^{0} && \text{Use the zero exponent rule}. \\ & = 1 \end{align}[/latex] |
4. [latex]\begin{align} \frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}& = 5{\left(r{s}^{2}\right)}^{2 - 2}&&\text{Use the quotient rule}. \\ & = 5{\left(r{s}^{2}\right)}^{0}&&\text{Use the zero exponent rule}. \\ & = 5\cdot 1 \\ & = 5 \end{align}[/latex] |
Try It 1
Simplify each expression using the zero exponent rule of exponents.
- [latex]\dfrac{{t}^{7}}{{t}^{7}}[/latex]
- [latex]\dfrac{{\left(d{e}^{2}\right)}^{11}}{2{\left(d{e}^{2}\right)}^{11}}[/latex]
- [latex]\dfrac{{w}^{4}\cdot {w}^{2}}{{w}^{6}}[/latex]
- [latex]\dfrac{-5{t}^{3}\cdot {t}^{4}}{{t}^{2}\cdot {t}^{5}}[/latex]
Negative Exponents
The quotient rule for exponents can also be used to determine what it means to have a negative exponent [latex]x^{-n}[/latex]. If [latex]m<n,\;m-n<0[/latex], so [latex]\dfrac{{a}^{m}}{{a}^{n}}={a}^{m-n}[/latex] will have a negative exponent.
Consider, for example, [latex]\dfrac{{x}^{2}}{{x}^{4}}={x}^{2-4}=x^{-2}[/latex]. We have a negative exponent, but what does that mean?
Another way to simplify [latex]\dfrac{{x}^{2}}{{x}^{4}}[/latex] is to expand the numerator and denominator then cancel common factors:
[latex]\begin{align}\frac{x^{2}}{x^{4}} &=\frac{x\cdot x}{x\cdot x\cdot x\cdot x} \\[1mm] &=\frac{\cancel{x}\cdot\cancel{x}}{\cancel{x}\cdot\cancel{x}\cdot x\cdot x} \\[1mm] & =\frac{1}{x^2}\end{align}[/latex]
Consequently, [latex]x^{-2}=\dfrac{1}{x^2}[/latex].
A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator. This can be generalized to [latex]a^{-n}=\dfrac{1}{a^n}[/latex].
If the negative exponent is on the denominator, [latex]\dfrac{1}{x^{-n}}[/latex], we can use division of fractions to simplify it:
[latex]\begin{aligned}\dfrac{1}{x^{-n}}&=1\div x^{-n}\\&=1\div \dfrac{1}{x^n}\\&=1\times \dfrac{x^n}{1}\\&=x^n\end{aligned}[/latex]
NEGATIVE EXPONENTS
For any real numbers [latex]a\neq0[/latex] and [latex]n[/latex],
[latex]a^{-n}=\dfrac{1}{a^n}[/latex]
and
[latex]\dfrac{1}{{a}^{-n}}=a^n[/latex]
A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar.
Example 2
Simplify the expressions. Write answers with positive exponents.
- [latex]\dfrac{x^3}{x^{10}}[/latex]
- [latex]\dfrac{z^2\cdot z}{z^4}[/latex]
- [latex]\dfrac{{\left(-5t^3\right)}^4}{\left(-5t^3\right)^8}[/latex]
Solution
[latex]1.\\ \begin{aligned}\dfrac{x^3}{x^{10}}&=x^{3 - 10}&&\text{Quotient rule}\\&=x^{-7}&&\text{Negative exponent rule}\\&=\dfrac{1}{x^7}\end{aligned}[/latex] | [latex]2.\\ \begin{aligned}\dfrac{z^2\cdot z}{z^4}&=\dfrac{z^{2+1}}{z^4}&&\text{Product rule}\\&=\dfrac{z^3}{z^4}&&\text{Quotient rule}\\&={z}^{3 - 4}\\&={z}^{-1}&&\text{Negative exponent rule}\\&=\dfrac{1}{z}\end{aligned}[/latex] |
[latex]3.\\ \begin{aligned}\dfrac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}&={\left(-5{t}^{3}\right)}^{4 - 8}&&\text{Product rule: the base is }-5t^3\\&={\left(-5{t}^{3}\right)}^{-4}&&\text{Negative exponent rule}\\&=\dfrac{1}{{\left(-5{t}^{3}\right)}^{4}}\end{aligned}[/latex] |
Try It 2
Simplify the expressions. Write answers with positive exponents.
- [latex]\dfrac{{\left(-3t\right)}^{2}}{{\left(-3t\right)}^{8}}[/latex]
- [latex]\dfrac{{f}^{47}}{{f}^{49}\cdot f}[/latex]
- [latex]\dfrac{2{k}^{4}}{5{k}^{7}}[/latex]
- [latex]\dfrac{\left(-3x^4\right)^5}{\left(-3x^4\right)^{12}}[/latex]
- [latex]\dfrac{5y^{-8}}{y^{-6}}[/latex]
Example 3
Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
- [latex]{b}^{2}\cdot {b}^{-8}[/latex]
- [latex]{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}[/latex]
- [latex]\dfrac{-7z}{{\left(-7z\right)}^{5}}[/latex]
Solution
1.
[latex]\begin{aligned}{b}^{2}\cdot {b}^{-8}&={b}^{2 + (-8)}&&\text{Product rule}\\&={b}^{-6}\\&=\frac{1}{{b}^{6}}&&\text{Negative exponent rule}\end{aligned}[/latex] |
2.
[latex]\begin{aligned}{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}&={\left(-x\right)}^{5+(-5)}&&\text{Product rule}\\&={\left(-x\right)}^{0}\\&=1&&\text{Zero exponent rule}\end{aligned}[/latex] |
3.
[latex]\begin{aligned}\dfrac{-7z}{{\left(-7z\right)}^{5}}&=\dfrac{{\left(-7z\right)}^{1}}{{\left(-7z\right)}^{5}}\\&={\left(-7z\right)}^{1 - 5}&&\text{Quotient rule}\\&={\left(-7z\right)}^{-4}\\&=\dfrac{1}{{\left(-7z\right)}^{4}}&&\text{Negative exponent rule}\end{aligned}[/latex] |
Try It 3
Simplify. Write answers with positive exponents.
- [latex]{t}^{-11}\cdot {t}^{6}[/latex]
- [latex]\dfrac{{25}^{12}}{{25}^{13}}[/latex]
The Power of a Product Rule
To simplify the power of a product of two exponential expressions, we can use the power of a product rule of exponents, which breaks up the power of a product of factors into the product of the powers of the factors. For example, consider [latex]{\left(pq\right)}^{3}[/latex]. We begin by using the associative and commutative properties of multiplication to regroup the factors:
In other words, [latex]{\left(pq\right)}^{3}={p}^{3}\cdot {q}^{3}[/latex].
The Power of a Product Rule
For any real numbers [latex]a,\;b[/latex] and [latex]n[/latex], the power of a product rule of exponents states that
Example 4
Simplify each of the following products as much as possible. Write answers with positive exponents.
- [latex]{\left(a{b}^{2}\right)}^{3}[/latex]
- [latex]{\left(2t\right)}^{15}[/latex]
- [latex]{\left(-2{w}^{3}\right)}^{3}[/latex]
- [latex]\dfrac{1}{{\left(-7z\right)}^{4}}[/latex]
- [latex]{\left({e}^{-2}{f}^{2}\right)}^{7}[/latex]
Solution
We can use the product and quotient rules and the new definitions to simplify each expression. If a number is raised to a power, we can evaluate it.
1.
[latex]\begin{aligned}{\left(a{b}^{2}\right)}^{3}&={\left(a\right)}^{3}\cdot {\left({b}^{2}\right)}^{3}&&\text{Power of a product rule}\\&={a}^{1\cdot 3}\cdot {b}^{2\cdot 3}\\&={a}^{3}{b}^{6}\end{aligned}[/latex] |
2.
[latex]\begin{aligned}(2{t})^{15}&={\left(2\right)}^{15}\cdot {\left(t\right)}^{15}&&\text{Power of a product rule}\\&={2}^{15}{t}^{15}&&\text{Evaluate }2^{15}\text{ using a calculator}\\&=32,768{t}^{15}\end{aligned}[/latex] |
3.
[latex]\begin{aligned}{\left(-2{w}^{3}\right)}^{3}&={\left(-2\right)}^{3}\cdot {\left({w}^{3}\right)}^{3}&&\text{Power of a product rule}\\&=-8\cdot {w}^{3\cdot 3}&&{(-2)}^3=-8\\&=-8{w}^{9}\end{aligned}[/latex] |
4.
[latex]\begin{aligned}\dfrac{1}{{\left(-7z\right)}^{4}}&=\dfrac{1}{{\left(-7\right)}^{4}\cdot {\left(z\right)}^{4}}&&\text{Power of a product rule}\\&=\dfrac{1}{2,401{z}^{4}}&&{(-7)}^4\text{ is evaluated}\end{aligned}[/latex] |
5.
[latex]\begin{aligned}{\left({e}^{-2}{f}^{2}\right)}^{7}&={\left({e}^{-2}\right)}^{7}\cdot {\left({f}^{2}\right)}^{7}&&\text{Power of a product rule}\\&={e}^{-2\cdot 7}\cdot {f}^{2\cdot 7}\\&={e}^{-14}{f}^{14}&&\text{Negative exponent rule}\\&=\dfrac{{f}^{14}}{{e}^{14}}\end{aligned}[/latex] |
Try It 4
Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
- [latex]{\left({g}^{2}{h}^{3}\right)}^{5}[/latex]
- [latex]{\left(5t\right)}^{3}[/latex]
- [latex]{\left(-3{y}^{5}\right)}^{3}[/latex]
- [latex]\dfrac{1}{{\left({a}^{6}{b}^{7}\right)}^{3}}[/latex]
- [latex]{\left({r}^{3}{s}^{-2}\right)}^{4}[/latex]
The Power of a Quotient Rule
To simplify the power of a quotient of two expressions, we can use the power of a quotient rule, which states that the power of a quotient of factors is the quotient of the powers of the factors.The power of a quotient rule is an extension of the power of a product rule since a quotient can be written as a product:
[latex]\begin{aligned}\left(\dfrac{a}{b}\right)^n&=\left(a\times\dfrac{1}{b}\right)^n\\&=a^n\times \left(b^{-1}\right)^n\\&=a^n\times b^{-n}\\&=a^n\times \dfrac{1}{b^n}\\&=\dfrac{a^n}{b^n}\end{aligned}[/latex].
The Power of a Quotient Rule
For any real numbers [latex]a,\;b[/latex] and [latex]n[/latex], provided [latex]b\neq0[/latex], the power of a quotient rule of exponents states that
Example 5
Simplify each of the following quotients as much as possible. Write answers with positive exponents.
- [latex]{\left(\dfrac{4}{{z}^{11}}\right)}^{3}[/latex]
- [latex]{\left(\dfrac{p}{{q}^{3}}\right)}^{6}[/latex]
- [latex]{\left(\dfrac{-1}{{t}^{2}}\right)}^{27}[/latex]
- [latex]{\left({j}^{3}{k}^{-2}\right)}^{4}[/latex]
- [latex]{\left({m}^{-2}{n}^{-2}\right)}^{3}[/latex]
Solution
1.
[latex]\begin{aligned}{\left(\dfrac{4}{{z}^{11}}\right)}^{3}&=\dfrac{{\left(4\right)}^{3}}{{\left({z}^{11}\right)}^{3}}&&\text{Quotient to a power rule}\\&=\dfrac{64}{{z}^{11\cdot 3}}&&\text{Evaluate }4^3=64\text{. Power to a power rule.} \\&=\dfrac{64}{{z}^{33}}\end{aligned}[/latex] |
2.
[latex]\begin{aligned}{\left(\dfrac{p}{{q}^{3}}\right)}^{6}&=\dfrac{{\left(p\right)}^{6}}{{\left({q}^{3}\right)}^{6}}&&\text{Power of a quotient rule}\\&=\dfrac{{p}^{1\cdot 6}}{{q}^{3\cdot 6}}&&\text{Power to a power rule}\\&=\dfrac{{p}^{6}}{{q}^{18}}\end{aligned}[/latex] |
3.
[latex]\begin{aligned}{\left(\dfrac{-1}{{t}^{2}}\right)}^{27}&=\dfrac{{\left(-1\right)}^{27}}{{\left({t}^{2}\right)}^{27}}&&\text{Power of a quotient rule}\\&=\dfrac{-1}{{t}^{2\cdot 27}}&&\text{Power of a power rule}\\&=\dfrac{-1}{{t}^{54}}&&\text{Put the negative sign in front of the fraction}\\&=-\dfrac{1}{{t}^{54}}\end{aligned}[/latex] |
4.
[latex]\begin{aligned}{\left({j}^{3}{k}^{-2}\right)}^{4}&={\left(\dfrac{{j}^{3}}{{k}^{2}}\right)}^{4}&&\text{Negative exponent rule}\\&=\dfrac{{\left({j}^{3}\right)}^{4}}{{\left({k}^{2}\right)}^{4}}&&\text{Power of a quotient rule}\\&=\dfrac{{j}^{3\cdot 4}}{{k}^{2\cdot 4}}&&\text{Power to a power rule}\\&=\dfrac{{j}^{12}}{{k}^{8}}\end{aligned}[/latex] |
5.
[latex]\begin{aligned}{\left({m}^{-2}{n}^{-2}\right)}^{3}&={\left(\dfrac{1}{{m}^{2}{n}^{2}}\right)}^{3}&&\text{Negative exponent rule}\\&=\dfrac{{\left(1\right)}^{3}}{{\left({m}^{2}{n}^{2}\right)}^{3}}&&\text{Power of a quotient rule}\\&=\dfrac{1}{{\left({m}^{2}\right)}^{3}{\left({n}^{2}\right)}^{3}}&&\text{Evaluate }1^3=1\\&=\dfrac{1}{{m}^{2\cdot 3}\cdot {n}^{2\cdot 3}}&&\text{Power to a power rule}\\&=\dfrac{1}{{m}^{6}{n}^{6}}\end{aligned}[/latex] |
Try It 5
Simplify each of the following quotients as much as possible. Write answers with positive exponents.
- [latex]{\left(\dfrac{{b}^{5}}{c}\right)}^{3}[/latex]
- [latex]{\left(\dfrac{5}{{u}^{8}}\right)}^{4}[/latex]
- [latex]{\left(\dfrac{-1}{{w}^{3}}\right)}^{35}[/latex]
- [latex]{\left({p}^{-4}{q}^{3}\right)}^{8}[/latex]
- [latex]{\left({c}^{-5}{d}^{-3}\right)}^{4}[/latex]
Simplifying Exponential Expressions
Recall that to simplify an expression means to rewrite it by combining terms or exponents. Evaluating an expression means to get a numerical answer. The rules for exponents can be combined to simplify expressions.
Example 6
Simplify each expression and write the answer with positive exponents only.
- [latex]{\left(6{m}^{2}{n}^{-1}\right)}^{3}[/latex]
- [latex]{17}^{5}\cdot {17}^{-4}\cdot {17}^{-3}[/latex]
- [latex]{\left(\dfrac{{u}^{-1}v}{{v}^{-1}}\right)}^{2}[/latex]
- [latex]\left(-2{a}^{3}{b}^{-1}\right)\left(5{a}^{-2}{b}^{2}\right)[/latex]
- [latex]{\left({x}^{2}\sqrt{2}\right)}^{4}{\left({x}^{2}\sqrt{2}\right)}^{-4}[/latex]
- [latex]\dfrac{{\left(3{w}^{2}\right)}^{5}}{{\left(6{w}^{-2}\right)}^{2}}[/latex]
Solution
1. [latex]\begin{align} {\left(6{m}^{2}{n}^{-1}\right)}^{3}& = {\left(6\right)}^{3}{\left({m}^{2}\right)}^{3}{\left({n}^{-1}\right)}^{3}&& \text{Power of a product rule} \\ & = {6}^{3}{m}^{2\cdot 3}{n}^{-1\cdot 3}&& \text{Power rule} \\ & = 216{m}^{6}{n}^{-3}&&\text{Evaluate: }2^6=216\text{ and simplify}. \\ & = \frac{216{m}^{6}}{{n}^{3}}&& \text{Negative exponent rule} \end{align}[/latex] |
2. [latex]\begin{align} {17}^{5}\cdot {17}^{-4}\cdot {17}^{-3}&=17^{5 +(-4)+(-3)}&& \text{Product rule} \\ & = {17}^{-2}&&\text{Negative exponent rule}\\&=\dfrac{1}{{17}^{2}}&&\text{Evaluate}\\&=\dfrac{1}{289}\end{align}[/latex] |
3. [latex]\begin{align} {\left(\dfrac{{u}^{-1}v}{{v}^{-1}}\right)}^{2}&=\dfrac{{\left({u}^{-1}v\right)}^{2}}{{\left({v}^{-1}\right)}^{2}}&&\text{Power of a quotient rule}\\&=\dfrac{{u}^{-2}{v}^{2}}{{v}^{-2}}&&\text{Power of a product rule}\\&={u}^{-2}{v}^{2-(-2)}&&\text{Quotient rule}\\&={u}^{-2}{v}^{4}&&\text{Evaluate: }2-(-2)=2+2=4\\&=\dfrac{{v}^{4}}{{u}^{2}}&&\text{Negative exponent rule}\end{align}[/latex] |
4. [latex]\begin{align}\left(-2a^3b^{-1}\right)\left(5a^{-2}b^2 \right)&=\left(-2\cdot 5\right)\left({a}^{3}\cdot {a}^{-2}\right)\left( {b}^{-1}\cdot {b}^{2}\right)&& \text{Commutative/associative properties} \\ & = -10{a}^{3+(-2)}{b}^{-1+2}&& \text{Product rule} \\ & = -10a^1b^1\\&=-10ab \end{align}[/latex] |
5. [latex]\begin{align} {\left({x}^{2}\sqrt{2}\right)}^{4}{\left({x}^{2}\sqrt{2}\right)}^{-4}& = {\left({x}^{2}\sqrt{2}\right)}^{4+(-4)} && \text{Product rule: base is }(x^2\sqrt{2}) \\ & = {\left({x}^{2}\sqrt{2}\right)}^{0}&& \text{Zero exponent rule}\\ & = 1 \end{align}[/latex] |
6. [latex]\begin{align} \frac{{\left(3{w}^{2}\right)}^{5}}{{\left(6{w}^{-2}\right)}^{2}}& = \frac{{\left(3\right)}^{5}\cdot {\left({w}^{2}\right)}^{5}}{{\left(6\right)}^{2}\cdot {\left({w}^{-2}\right)}^{2}}&& \text{Power of a product rule} \\ & = \frac{{3}^{5}{w}^{2\cdot 5}}{{6}^{2}{w}^{-2\cdot 2}}&& \text{Power rule} \\ & = \frac{243{w}^{10}}{36{w}^{-4}} && \text{Evaluate: }3^5=243\text{ and }6^2=36 \\ & = \frac{27{w}^{10-\left(-4\right)}}{4}&& \text{Quotient rule and simplify fraction} \\ & = \frac{27{w}^{14}}{4}\end{align}[/latex] |
Try It 6
Simplify each expression and write the answer with positive exponents only.
- [latex]{\left(2u{v}^{-2}\right)}^{-3}[/latex]
- [latex]{x}^{8}\cdot {x}^{-12}\cdot x[/latex]
- [latex]{\left(\dfrac{{e}^{2}{f}^{-3}}{{f}^{-1}}\right)}^{2}[/latex]
- [latex]\left(9{r}^{-5}{s}^{3}\right)\left(3{r}^{6}{s}^{-4}\right)[/latex]
- [latex]{\left(\dfrac{4}{9}t{w}^{-2}\right)}^{-3}{\left(\dfrac{4}{9}t{w}^{-2}\right)}^{3}[/latex]
- [latex]\dfrac{{\left(2{h}^{2}k\right)}^{4}}{{\left(7{h}^{-1}{k}^{2}\right)}^{2}}[/latex]
Simplifying Exponential Functions
Each of the properties of exponents can be used to simplify and write equivalent exponential functions. For example, the function [latex]f(x)=2^{x+3}[/latex] can be written as the equivalent exponential function [latex]f(x)=8\left(2^x\right)[/latex]:
[latex]\begin{aligned}f(x)&=2^{x+3}\\&=2^x\cdot 2^3\\&=2^x\cdot 8\\&=8\left(2^x\right)\end{aligned}[/latex]
Being able to simplify an exponential function into the standard form [latex]f(x)=ar^{x-h}+k[/latex] makes the function easier to graph and allows us to determine the transformations that were made from the parent function [latex]f(x)=r^x[/latex]. In the case of [latex]f(x)=2^{x+3}=8\left(2^x\right)[/latex], the 8 tells us that the graph of [latex]f(x)=2^x[/latex] has been stretched by a factor of 8.
Example 7
Use exponential rules to write equivalent exponential functions.
- [latex]f(x)=3^{x+2}[/latex]
- [latex]g(x)=5^{2x+1}[/latex]
- [latex]h(x)=4^{3x-2}[/latex]
Solution
1.
[latex]\begin{aligned}f(x)&=3^{x+2}\\&=3^x\cdot 3^2&&\text{Product rule (in reverse)}\\&=3^x\cdot 9&&\text{Evaluate }3^2=9\\&=9\left(3^x\right)&&\text{Write in standard form}\end{aligned}[/latex]
2.
[latex]\begin{aligned}g(x)&=5^{2x+1}\\&=5^{2x}\cdot 5^1&&\text{Product rule (in reverse)}\\&=\left(5^2\right)^x\cdot 5&&\text{Evaluate }5^1=5\\&=5\left(25^x\right)&&\text{Write in standard form}\end{aligned}[/latex]
3.
[latex]\begin{aligned}h(x)&=4^{3x-2}\\&=4^{3x}\cdot 4^{-2}&&\text{Product rule (in reverse)}\\&=\left(4^3\right)^x\cdot \dfrac{1}{4^2}&&\text{Power to a power rule (in reverse) and negative exponent rule}\\&=64^x\cdot \dfrac{1}{16}&&\text{Evaluate: }4^3=64\text{ and }4^2=16\\&=\dfrac{1}{16}\left(64^x\right)&&\text{Write in standard form}\end{aligned}[/latex]
Try It 11
Use exponential rules to write equivalent exponential functions.
- [latex]f(x)=2^{x+4}[/latex]
- [latex]g(x)=3^{x-2}[/latex]
- [latex]h(x)=5^{2x-1}[/latex]