0.1 Exponent Rules

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We explore the rules for exponents.

Understanding Exponential Expressions

An exponential expression takes the form $a^b$ , where $a$ is the base and $b$ is the exponent. For instance, $2^3, x^n$ and $e^x$ are exponential expressions. When we encounter the expression $a^b$ , where $b$ is a positive integer, it signifies the act of multiplying $a$ by itself $b$ times. Visually, this can be expressed as: $a^b = \underbrace {a \cdot a \cdot \ldots \cdot a}_\text {$b$ times}$

(problem 1) Compute the value of each of the following exponential expressions (without a calculator): $a) \;6^2 = \answer {36} \quad b) \;7^2 = \answer {49} \quad c) \;8^2 = \answer {64} \quad d) \;9^2 = \answer {81}$

(problem 2) Compute the value of each of the following exponential expressions (without a calculator): $a) \ 2^3 = \answer {8} \; \ b) \ 3^3 = \answer {27} \; \ c) \ 4^3 = \answer {64} \; \ d) \ 5^3 = \answer {125} \; \ e) \ 10^3 = \answer {1000}$

(problem 3) Compute the value of each of the following exponential expressions (without a calculator): $a) \ 2^4 = \answer {16} \; \ b) \ 3^4 = \answer {81} \; \ c) \ 5^4 = \answer {625} \; \ d) \ 10^4 = \answer {10000}$

(problem 4) Compute the value of each of the following exponential expressions (without a calculator): $a) \;2^5 = \answer {32} \quad b) \;2^6 = \answer {64} \quad c) \;2^7 = \answer {128} \quad d) \;2^8 = \answer {256}$

Note that $64 = 8^2 = 4^3 = 2^6$ .

According to the order of operations, exponents are computed before other multiplications and divisions. Hence, if $f(x) = 5x^2$ then $f(3) = 5(3^2) = 5(9) = 45$ .

(problem 5) Evaluate the function $f(x) = 4x^2$ as indicated (without a calculator): $a) \ f(2) = \answer {16} \; \ b) \ f(-2) = \answer {16} \; \ c) \ f(-3) = \answer {36} \; \ d) \ f(-5) = \answer {100}$

(problem 6) Compute the value of each of the following exponential expressions (without a calculator): $a) \ -2^2 = \answer {-4} \; \ b) \ (-2)^2 = \answer {4} \; \ c) \ -1^4 = \answer {-1} \; \ d) \ (-1)^4 = \answer {1}$

Adding Exponents

When multiplying exponential expressions with the same base, we add the exponents: $a^b \cdot a^c = a^{b+c}$ For positive integer exponents, the rationale is shown below: $a^b \cdot a^c = (\underbrace {a \cdot a \cdot a \cdot \ldots \cdot a}_\text {$b$ times}) \cdot (\underbrace {a \cdot a \cdot a \cdot \ldots \cdot a}_\text {$c$ times})$

$= \underbrace {a \cdot a \cdot a \cdot \ldots \cdot a}_\text {$b+c$ times}$

$= a^{b+c}$

(problem 7) Combine each of the following exponential expressions into a single exponential expression: $a) \ a^4\cdot a^5 = \answer {a^9} \; \ b) \ r^2 \cdot r^3 = \answer {r^5} \; \ c) \ n\cdot n^6 = \answer {n^7} \; \ d) \ x^{14}\cdot x^{17} = \answer {x^{31}}$

(problem 8) Distribute: $x^2(x^4 + x^7) = \answer {x^9+x^6}$

Subtracting Exponents

When dividing exponential expressions with the same base, we subtract the exponents: $\frac {a^b}{a^c} = a^{b-c}$ For positive integer exponents $b$ and $c$ with $b > c$ , the rationale is shown below:

\begin{eqnarray*} \frac{a^b}{a^c} &=& \frac{\overbrace{a \cdot a \cdot a \cdot \ldots \cdot a}^\text{$b$ times}}{\underbrace{ a \cdot a \cdot \ldots \cdot a}_\text{$c$ times}}\\[10pt] &=& \underbrace{ a \cdot a \cdot \ldots \cdot a}_\text{$b-c$ times}\\[10pt] &=& a^{b-c} \end{eqnarray*}

(problem 9) Combine each of the following exponential expressions into a single exponential expression: $a) \ \frac {a^2}{a} = \answer {a} \; \ b) \ \frac {r^3}{r} = \answer {r^2} \; \ c) \ \frac {n^5}{n^3} = \answer {n^2} \; \ d) \ \frac {x^7}{x^3} = \answer {x^4}$

(problem 10) Divide: $\frac {x^8 + x^5 + x^2}{x^2} = \answer {x^6 + x^3 + 1}$

The Zero Exponent

For equal bases $a$ , $\frac {a^b}{a^b} = 1$ Also, using the exponent subtraction rule, $\frac {a^b}{a^b} = a^{b-b} = a^0$ Therefore, we define $a^0 = 1$ for $a \neq 0$ .

If $a \neq 0$ , then we define $a^0 \equiv 1$ .

(problem 11) Evaluate each of the following exponential expressions (without a calculator): $a) \ -3^0 = \answer {-1} \; \ b) \ e^0 = \answer {1} \; \ c) \ \pi ^0 = \answer {1} \; \ d) \ (-1)^0 = \answer {1}$

Negative Exponents

If the exponent in the denominator is greater than the exponent in the numerator, then subtracting exponents yields a negative exponent. Consider the following example:

$\frac {a^3}{a^5} = a^{3-5} = a^{-2}$ On the other hand, we can rewrite this ratio of exponential expressions as follows $\frac {a^3}{a^5} = \frac {a\cdot a\cdot a}{a\cdot a\cdot a \cdot a\cdot a} = \frac {1}{a\cdot a} = \frac {1}{a^2}$ Putting these together yields $a^{-2} = \frac {1}{a^2}$ We generalize this idea and define negative exponents as follows:

For $a \neq 0$ and $n$ a positive integer, we define $a^{-n}$ by $a^{-n} = \frac {1}{a^n}$

(problem 12) Rewrite the following expressions using negative exponents: $a) \ \frac {3}{x^5} = \answer {3x^{-5}} \; \ b) \ \frac {1}{3x^5} = \answer {x^{-5}/3} \; \ c) \ \frac {1}{u^2} = \answer {u^{-2}}$

(problem 13) Simply the following expression using negative exponents: $\frac {x^4 + x^2 + 1}{x^5} = \answer {x^{-1} + x^{-3} + x^{-5}}$

Multiplying Exponents

When raising an exponential expression to a power, we multiply the exponents. In the justification presented is below, the exponents, $b$ and $c$ , are positive integers: $\left (a^b\right )^c = \underbrace {a^b \cdot a^b \cdot \ldots \cdot a^b}_\text {$c$ times}$ Adding the exponents gives: $= a^{\overbrace {b + b + \dots +b}^\text {$c$ times}} = a^{bc}$

(problem 14) Combine each of the following into a single exponential expression (without a calculator): $a) \ \left (a^4\right )^5 = \answer {a^{20}} \; \ b) \ \left (r^2\right )^3 = \answer {r^6} \; \ c) \ \left (n^6\right )^6 = \answer {n^{36}} \; \ d) \ \left (x^{12}\right )^{12} = \answer {x^{144}}$

Rational Exponents

Let $n$ be a positive integer and let $a\geq 0$ if $n$ is even. Then $\sqrt [n] a$ is the number which when multiplied by itself $n$ times gives $a$ . In other words: $(\sqrt [n] a)^n = a$ Allow for a moment the use of the exponent $\frac {1}{n}$ . Using the rule for multiplying exponents, we have $\left (a^\frac {1}{n}\right )^n = a^{\left (\frac {1}{n} \cdot n\right )} = a^1 = a$ Thus, $\sqrt [n] a$ and $a^\frac {1}{n}$ both yield $a$ when raised to the $n^\text {th}$ power. Hence, we define $a^{1/n} = \sqrt [n] a$ .

Let $n$ be a positive integer. We define $a^{\frac {1}{n}} \equiv \sqrt [n] a$ where $a \geq 0$ if $n$ is even.

Now consider $(\sqrt [n] a)^m = \left (a^{1/n}\right )^m = a^{m/n}$ This leads to the following definition.

Let $m$ and $n$ be positive integers. Then we define $a^{\frac {m}{n}} \equiv (\sqrt [n] a)^m = \sqrt [n]{a^m}$ where $a \geq 0$ if $n$ is even.

(problem 15) Rewrite the following expressions using rational (and possibly negative) exponents: $a) \ \frac {1}{\sqrt a} = \answer {a^{-1/2}} \; \ b) \;\sqrt [3] r = \answer {r^{1/3}} \; \ c) \ \sqrt [4]{u^3} = \answer {u^{3/4}}$

(problem 16) Simplify into a sum of powers of $x$ : $\frac {x^5 + \sqrt [3]{x^2} + x^{-3}}{x\sqrt x} = \answer {x^{7/2} + x^{-5/6} + x^{-9/2}}$

2024-09-27 13:58:22