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Exponents and Scientific Notation

Use the product rule of exponents.

Using the Product Rule of Exponents

Consider the product ${x}^{3}⋅{x}^{4}.$ Both terms have the same base, x, but they are raised to different exponents. Expand each expression, and then rewrite the resulting expression.

\[\begin{array}{lll}{x}^{3}⋅{x}^{4} & = & \overset{3\ \text{factors}}{\overset{}{x⋅x⋅x}}⋅\overset{4\ \text{factors}}{\overset{}{x⋅x⋅x⋅x}} \\ & = & \overset{7\ \text{factors}}{\overset{}{x⋅x⋅x⋅x⋅x⋅x⋅x}} \\ & = & {x}^{7}\end{array}\]

The result is that ${x}^{3}⋅{x}^{4}={x}^{3+4}={x}^{7}.$

Notice that the exponent of the product is the sum of the exponents of the terms. In other words, when multiplying exponential expressions with the same base, we write the result with the common base and add the exponents. This is the product rule of exponents.

\[{a}^{m}⋅{a}^{n}={a}^{m+n}\]

Now consider an example with real numbers.

\[{2}^{3}⋅{2}^{4}={2}^{3+4}={2}^{7}\]

We can always check that this is true by simplifying each exponential expression. We find that ${2}^{3}$ is 8, ${2}^{4}$ is 16, and ${2}^{7}$ is 128. The product $8⋅16$ equals 128, so the relationship is true. We can use the product rule of exponents to simplify expressions that are a product of two numbers or expressions with the same base but different exponents.

Example

Try it.

Write each of the following products with a single base. Do not simplify further.

ⓐ ${t}^{5}⋅{t}^{3}$
ⓑ ${(-3)}^{5}⋅(-3)$
ⓒ ${x}^{2}⋅{x}^{5}⋅{x}^{3}$

Solution

Use the product rule to simplify each expression.

ⓐ ${t}^{5}⋅{t}^{3}={t}^{5+3}={t}^{8}$
ⓑ ${(-3)}^{5}⋅(-3)={(-3)}^{5}⋅{(-3)}^{1}={(-3)}^{5+1}={(-3)}^{6}$
ⓒ ${x}^{2}⋅{x}^{5}⋅{x}^{3}$

At first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two.

\[{x}^{2}⋅{x}^{5}⋅{x}^{3}=({x}^{2}⋅{x}^{5})⋅{x}^{3}=({x}^{2+5})⋅{x}^{3}={x}^{7}⋅{x}^{3}={x}^{7+3}={x}^{10}\]

Notice we get the same result by adding the three exponents in one step.

\[{x}^{2}⋅{x}^{5}⋅{x}^{3}={x}^{2+5+3}={x}^{10}\]

Using the Quotient Rule of Exponents

The quotient rule of exponents allows us to simplify an expression that divides two numbers with the same base but different exponents. In a similar way to the product rule, we can simplify an expression such as $\frac{{y}^{m}}{{y}^{n}},$ where $m>n.$ Consider the example $\frac{{y}^{9}}{{y}^{5}}.$ Perform the division by canceling common factors.

\[\begin{array}{lll}\frac{{y}^{9}}{{y}^{5}} & = & \frac{y⋅y⋅y⋅y⋅y⋅y⋅y⋅y⋅y}{y⋅y⋅y⋅y⋅y} \\ & = & \frac{y⋅y⋅y⋅y⋅y⋅y⋅y⋅y⋅y}{y⋅y⋅y⋅y⋅y} \\ & = & \frac{y⋅y⋅y⋅y}{1} \\ & = & {y}^{4}\end{array}\]

Notice that the exponent of the quotient is the difference between the exponents of the divisor and dividend.

\[\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}\]

In other words, when dividing exponential expressions with the same base, we write the result with the common base and subtract the exponents.

\[\frac{{y}^{9}}{{y}^{5}}={y}^{9-5}={y}^{4}\]

For the time being, we must be aware of the condition $m>n.$ Otherwise, the difference $m-n$ could be zero or negative. Those possibilities will be explored shortly. Also, instead of qualifying variables as nonzero each time, we will simplify matters and assume from here on that all variables represent nonzero real numbers.

Example

Try it.

Write each of the following products with a single base. Do not simplify further.

ⓐ $\frac{{(-2)}^{14}}{{(-2)}^{9}}$
ⓑ $\frac{{t}^{23}}{{t}^{15}}$
ⓒ $\frac{{(z\sqrt{2})}^{5}}{z\sqrt{2}}$

Solution

Use the quotient rule to simplify each expression.

ⓐ $\frac{{(-2)}^{14}}{{(-2)}^{9}}={(-2)}^{14-9}={(-2)}^{5}$
ⓑ $\frac{{t}^{23}}{{t}^{15}}={t}^{23-15}={t}^{8}$
ⓒ $\frac{{(z\sqrt{2})}^{5}}{z\sqrt{2}}={(z\sqrt{2})}^{5-1}={(z\sqrt{2})}^{4}$

Using the Power Rule of Exponents

Suppose an exponential expression is raised to some power. Can we simplify the result? Yes. To do this, we use the power rule of exponents. Consider the expression ${({x}^{2})}^{3}.$ The expression inside the parentheses is multiplied twice because it has an exponent of 2. Then the result is multiplied three times because the entire expression has an exponent of 3.

\[\begin{array}{lll}{({x}^{2})}^{3} & = & \overset{3\ \text{factors}}{\overset{}{({x}^{2})⋅({x}^{2})⋅({x}^{2})}} \\ & = & \overset{3\ \text{factors}}{\overset{}{(\overset{2\ \text{factors}}{\overset{︷}{x⋅x}})⋅(\overset{2\ \text{factors}}{\overset{︷}{x⋅x}})⋅(\overset{2\ \text{factors}}{\overset{︷}{x⋅x}})}} \\ & = & x⋅x⋅x⋅x⋅x⋅x \\ & = & {x}^{6}\end{array}\]

The exponent of the answer is the product of the exponents: ${({x}^{2})}^{3}={x}^{2⋅3}={x}^{6}.$ In other words, when raising an exponential expression to a power, we write the result with the common base and the product of the exponents.

\[{({a}^{m})}^{n}={a}^{m⋅n}\]

Be careful to distinguish between uses of the product rule and the power rule. When using the product rule, different terms with the same bases are raised to exponents. In this case, you add the exponents. When using the power rule, a term in exponential notation is raised to a power. In this case, you multiply the exponents.

\[\begin{array}{lllllllllll} & & \text{Product Rule} & & & & & \ \text{Power Rule} & & \\ {5}^{3}⋅{5}^{4} & = & {5}^{3+4} & = & {5}^{7} & \ \text{but}\ & \ {({5}^{3})}^{4} & = & {5}^{3⋅4} & = & {5}^{12} \\ {x}^{5}⋅{x}^{2} & = & {x}^{5+2} & = & {x}^{7} & \ \text{but}\ & {({x}^{5})}^{2} & = & {x}^{5⋅2} & = & {x}^{10} \\ {(3a)}^{7}⋅{(3a)}^{10} & = & {(3a)}^{7+10} & = & {(3a)}^{17} & \ \text{but}\ & {({(3a)}^{7})}^{10} & = & {(3a)}^{7⋅10} & = & {(3a)}^{70}\end{array}\]

Example

Try it.

Write each of the following products with a single base. Do not simplify further.

ⓐ ${({x}^{2})}^{7}$
ⓑ ${({(2t)}^{5})}^{3}$
ⓒ ${({(-3)}^{5})}^{11}$

Solution

Use the power rule to simplify each expression.

ⓐ ${({x}^{2})}^{7}={x}^{2⋅7}={x}^{14}$
ⓑ ${({(2t)}^{5})}^{3}={(2t)}^{5⋅3}={(2t)}^{15}$
ⓒ ${({(-3)}^{5})}^{11}={(-3)}^{5⋅11}={(-3)}^{55}$

Using the Zero Exponent Rule of Exponents

Return to the quotient rule. We made the condition that $m>n$ so that the difference $m-n$ would never be zero or negative. What would happen if $m=n?$ In this case, we would use the zero exponent rule of exponents to simplify the expression to 1. To see how this is done, let us begin with an example.

\[\frac{{t}^{8}}{{t}^{8}}=\frac{{t}^{8}}{{t}^{8}}=1\]

If we were to simplify the original expression using the quotient rule, we would have

\[\frac{{t}^{8}}{{t}^{8}}={t}^{8-8}={t}^{0}\]

If we equate the two answers, the result is ${t}^{0}=1.$ This is true for any nonzero real number, or any variable representing a real number.

\[{a}^{0}=1\]

The sole exception is the expression ${0}^{0}.$ This appears later in more advanced courses, but for now, we will consider the value to be undefined.

Example

Try it.

Simplify each expression using the zero exponent rule of exponents.

ⓐ $\frac{{c}^{3}}{{c}^{3}}$
ⓑ $\frac{-3{x}^{5}}{{x}^{5}}$
ⓒ $\frac{{({j}^{2}k)}^{4}}{({j}^{2}k)⋅{({j}^{2}k)}^{3}}$
ⓓ $\frac{5{(r{s}^{2})}^{2}}{{(r{s}^{2})}^{2}}$

Solution

Use the zero exponent and other rules to simplify each expression.

ⓐ
$\begin{array}{lll}\frac{{c}^{3}}{{c}^{3}} & = & {c}^{3-3} \\ & = & {c}^{0} \\ & = & 1\end{array}$

ⓑ
$\begin{array}{lll}\frac{-3{x}^{5}}{{x}^{5}} & = & -3⋅\frac{{x}^{5}}{{x}^{5}} \\ & = & -3⋅{x}^{5-5} \\ & = & -3⋅{x}^{0} \\ & = & -3⋅1 \\ & = & -3\end{array}$

ⓒ
$\begin{array}{llll}\frac{{({j}^{2}k)}^{4}}{({j}^{2}k)⋅{({j}^{2}k)}^{3}} & = & \frac{{({j}^{2}k)}^{4}}{{({j}^{2}k)}^{1+3}} & \ \text{Use the product rule in the denominator}. \\ & = & \frac{{({j}^{2}k)}^{4}}{{({j}^{2}k)}^{4}} & \ \text{Simplify}. \\ & = & {({j}^{2}k)}^{4-4} & \ \text{Use the quotient rule}. \\ & = & {({j}^{2}k)}^{0} & \ \text{Simplify}. \\ & = & 1 & \end{array}$

ⓓ
$\begin{array}{llll}\frac{5{(r{s}^{2})}^{2}}{{(r{s}^{2})}^{2}} & = & 5{(r{s}^{2})}^{2-2} & \ \text{Use the quotient rule}. \\ & = & 5{(r{s}^{2})}^{0} & \ \text{Simplify}. \\ & = & 5⋅1 & \ \text{Use the zero exponent rule}. \\ & = & 5 & \ \text{Simplify}.\end{array}$

Using the Negative Rule of Exponents

Another useful result occurs if we relax the condition that $m>n$ in the quotient rule even further. For example, can we simplify $\frac{{h}^{3}}{{h}^{5}}?$ When \(mnegative rule of exponents to simplify the expression to its reciprocal.

Divide one exponential expression by another with a larger exponent. Use our example, $\frac{{h}^{3}}{{h}^{5}}.$

\[\begin{array}{lll}\frac{{h}^{3}}{{h}^{5}} & = & \frac{h⋅h⋅h}{h⋅h⋅h⋅h⋅h} \\ & = & \frac{h⋅h⋅h}{h⋅h⋅h⋅h⋅h} \\ & = & \frac{1}{h⋅h} \\ & = & \frac{1}{{h}^{2}}\end{array}\]

If we were to simplify the original expression using the quotient rule, we would have

\[\begin{array}{lll}\frac{{h}^{3}}{{h}^{5}} & = & {h}^{3-5} \\ & = & \ {h}^{-2}\end{array}\]

Putting the answers together, we have ${h}^{-2}=\frac{1}{{h}^{2}}.$ This is true for any nonzero real number, or any variable representing a nonzero real number.

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 or vice versa.

\[\begin{array}{lll}{a}^{-n}=\frac{1}{{a}^{n}} & \text{and} & {a}^{n}=\frac{1}{{a}^{-n}}\end{array}\]

We have shown that the exponential expression ${a}^{n}$ is defined when $n$ is a natural number, 0, or the negative of a natural number. That means that ${a}^{n}$ is defined for any integer $n.$ Also, the product and quotient rules and all of the rules we will look at soon hold for any integer $n.$

Example

Try it.

Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.

ⓐ $\frac{{\theta }^{3}}{{\theta }^{10}}$
ⓑ $\frac{{z}^{2}⋅z}{{z}^{4}}$
ⓒ $\frac{{(-5{t}^{3})}^{4}}{{(-5{t}^{3})}^{8}}$

Solution

ⓐ $\frac{{\theta }^{3}}{{\theta }^{10}}={\theta }^{3-10}={\theta }^{-7}=\frac{1}{{\theta }^{7}}$
ⓑ $\frac{{z}^{2}⋅z}{{z}^{4}}=\frac{{z}^{2+1}}{{z}^{4}}=\frac{{z}^{3}}{{z}^{4}}={z}^{3-4}={z}^{-1}=\frac{1}{z}$
ⓒ $\frac{{(-5{t}^{3})}^{4}}{{(-5{t}^{3})}^{8}}={(-5{t}^{3})}^{4-8}={(-5{t}^{3})}^{-4}=\frac{1}{{(-5{t}^{3})}^{4}}$

Condensed — the full section is in OpenStax College Algebra 2e.

Finding the Power of a Product

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 instance, consider ${(pq)}^{3}.$ We begin by using the associative and commutative properties of multiplication to regroup the factors.

\[\begin{array}{lll}{(pq)}^{3} & = & \overset{3\ \text{factors}}{\overset{}{(pq)⋅(pq)⋅(pq)}} \\ & = & p⋅q⋅p⋅q⋅p⋅q \\ & = & \overset{3\ \text{factors}}{\overset{}{p⋅p⋅p}}⋅\overset{3\ \text{factors}}{\overset{}{q⋅q⋅q}} \\ & = & {p}^{3}⋅{q}^{3}\end{array}\]

In other words, ${(pq)}^{3}={p}^{3}⋅{q}^{3}.$

Example

Try it.

Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.

ⓐ ${(a{b}^{2})}^{3}$
ⓑ ${(2t)}^{15}$
ⓒ ${(-2{w}^{3})}^{3}$
ⓓ $\frac{1}{{(-7z)}^{4}}$
ⓔ ${({e}^{-2}{f}^{2})}^{7}$

Solution

Use the product and quotient rules and the new definitions to simplify each expression.

ⓐ ${(a{b}^{2})}^{3}={(a)}^{3}⋅{({b}^{2})}^{3}={a}^{1⋅3}⋅{b}^{2⋅3}={a}^{3}{b}^{6}$
ⓑ ${(2t)}^{15}={(2)}^{15}⋅{(t)}^{15}={2}^{15}{t}^{15}=32,768{t}^{15}$
ⓒ ${(-2{w}^{3})}^{3}={(-2)}^{3}⋅{({w}^{3})}^{3}=-8⋅{w}^{3⋅3}=-8{w}^{9}$
ⓓ $\frac{1}{{(-7z)}^{4}}=\frac{1}{{(-7)}^{4}⋅{(z)}^{4}}=\frac{1}{2,401{z}^{4}}$
ⓔ ${({e}^{-2}{f}^{2})}^{7}={({e}^{-2})}^{7}⋅{({f}^{2})}^{7}={e}^{-2⋅7}⋅{f}^{2⋅7}={e}^{-14}{f}^{14}=\frac{{f}^{14}}{{e}^{14}}$

Finding the Power of a Quotient

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. For example, let’s look at the following example.

\[{({e}^{-2}{f}^{2})}^{7}=\frac{{f}^{14}}{{e}^{14}}\]

Let’s rewrite the original problem differently and look at the result.

\[\begin{array}{lll}{({e}^{-2}{f}^{2})}^{7} & = & {(\frac{{f}^{2}}{{e}^{2}})}^{7} \\ & = & \frac{{f}^{14}}{{e}^{14}}\end{array}\]

It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule.

\[\begin{array}{lll}{({e}^{-2}{f}^{2})}^{7} & = & {(\frac{{f}^{2}}{{e}^{2}})}^{7} \\ & = & \frac{{({f}^{2})}^{7}}{{({e}^{2})}^{7}} \\ & = & \frac{{f}^{2⋅7}}{{e}^{2⋅7}} \\ & = & \frac{{f}^{14}}{{e}^{14}}\end{array}\]

Example

Try it.

Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.

ⓐ ${(\frac{4}{{z}^{11}})}^{3}$
ⓑ ${(\frac{p}{{q}^{3}})}^{6}$
ⓒ ${(\frac{-1}{{t}^{2}})}^{27}$
ⓓ ${({j}^{3}{k}^{-2})}^{4}$
ⓔ ${({m}^{-2}{n}^{-2})}^{3}$

Solution

ⓐ ${(\frac{4}{{z}^{11}})}^{3}=\frac{{(4)}^{3}}{{({z}^{11})}^{3}}=\frac{64}{{z}^{11⋅3}}=\frac{64}{{z}^{33}}$
ⓑ ${(\frac{p}{{q}^{3}})}^{6}=\frac{{(p)}^{6}}{{({q}^{3})}^{6}}=\frac{{p}^{1⋅6}}{{q}^{3⋅6}}=\frac{{p}^{6}}{{q}^{18}}$
ⓒ ${(\frac{-1}{{t}^{2}})}^{27}=\frac{{(-1)}^{27}}{{({t}^{2})}^{27}}=\frac{-1}{{t}^{2⋅27}}=\frac{-1}{{t}^{54}}=-\frac{1}{{t}^{54}}$
ⓓ ${({j}^{3}{k}^{-2})}^{4}={(\frac{{j}^{3}}{{k}^{2}})}^{4}=\frac{{({j}^{3})}^{4}}{{({k}^{2})}^{4}}=\frac{{j}^{3⋅4}}{{k}^{2⋅4}}=\frac{{j}^{12}}{{k}^{8}}$
ⓔ ${({m}^{-2}{n}^{-2})}^{3}={(\frac{1}{{m}^{2}{n}^{2}})}^{3}=\frac{{(1)}^{3}}{{({m}^{2}{n}^{2})}^{3}}=\frac{1}{{({m}^{2})}^{3}{({n}^{2})}^{3}}=\frac{1}{{m}^{2⋅3}⋅{n}^{2⋅3}}=\frac{1}{{m}^{6}{n}^{6}}$

Condensed — the full section is in OpenStax College Algebra 2e.

Simplifying Exponential Expressions

Recall that to simplify an expression means to rewrite it by combining terms or exponents; in other words, to write the expression more simply with fewer terms. The rules for exponents may be combined to simplify expressions.

Example

Try it.

Simplify each expression and write the answer with positive exponents only.

ⓐ ${(6{m}^{2}{n}^{-1})}^{3}$
ⓑ ${17}^{5}⋅{17}^{-4}⋅{17}^{-3}$
ⓒ ${(\frac{{u}^{-1}v}{{v}^{-1}})}^{2}$
ⓓ $(-2{a}^{3}{b}^{-1})(5{a}^{-2}{b}^{2})$
ⓔ ${({x}^{2}\sqrt{2})}^{4}{({x}^{2}\sqrt{2})}^{-4}$
ⓕ $\frac{{(3{w}^{2})}^{5}}{{(6{w}^{-2})}^{2}}$

Solution

ⓐ
$\begin{array}{llll}{(6{m}^{2}{n}^{-1})}^{3} & = & {(6)}^{3}{({m}^{2})}^{3}{({n}^{-1})}^{3} & \ \text{The power of a product rule} \\ & = & {6}^{3}{m}^{2⋅3}{n}^{-1⋅3} & \ \text{The power rule} \\ & = & \ 216{m}^{6}{n}^{-3} & \ \text{Simplify}. \\ & = & \frac{216{m}^{6}}{{n}^{3}} & \ \text{The negative exponent rule}\end{array}$
ⓑ
$\begin{array}{llll}{17}^{5}⋅{17}^{-4}⋅{17}^{-3} & = & {17}^{5-4-3} & \ \text{The product rule} \\ & = & {17}^{-2} & \ \text{Simplify}. \\ & = & \frac{1}{{17}^{2}}\ \text{or }\frac{1}{289} & \ \text{The negative exponent rule}\end{array}$
ⓒ
$\begin{array}{llll}{(\frac{{u}^{-1}v}{{v}^{-1}})}^{2} & = & \frac{{({u}^{-1}v)}^{2}}{{({v}^{-1})}^{2}} & \ \text{The power of a quotient rule} \\ & = & \frac{{u}^{-2}{v}^{2}}{{v}^{-2}} & \ \text{The power of a product rule} \\ & = & {u}^{-2}{v}^{2-(-2)} & \ \text{The quotient rule} \\ & = & {u}^{-2}{v}^{4} & \ \text{Simplify}. \\ & = & \frac{{v}^{4}}{{u}^{2}} & \ \text{The negative exponent rule}\end{array}$
ⓓ
$\begin{array}{llll}(-2{a}^{3}{b}^{-1})(5{a}^{-2}{b}^{2}) & = & -2⋅5⋅{a}^{3}⋅{a}^{-2}⋅{b}^{-1}⋅{b}^{2} & \ \text{Commutative and associative laws of multiplication} \\ & = & -10⋅{a}^{3-2}⋅{b}^{-1+2} & \ \text{The product rule} \\ & = & -10ab & \ \text{Simplify}.\end{array}$
ⓔ
$\begin{array}{llll}{({x}^{2}\sqrt{2})}^{4}{({x}^{2}\sqrt{2})}^{-4} & = & {({x}^{2}\sqrt{2})}^{4-4} & \ \text{The product rule} \\ & = & \ {({x}^{2}\sqrt{2})}^{0} & \ \text{Simplify}. \\ & = & 1 & \ \text{The zero exponent rule}\end{array}$
ⓕ
$\begin{array}{llll}\frac{{(3{w}^{2})}^{5}}{{(6{w}^{-2})}^{2}} & = & \frac{{(3)}^{5}⋅{({w}^{2})}^{5}}{{(6)}^{2}⋅{({w}^{-2})}^{2}} & \ \text{The power of a product rule} \\ & = & \frac{{3}^{5}{w}^{2⋅5}}{{6}^{2}{w}^{-2⋅2}} & \ \text{The power rule} \\ & = & \frac{243{w}^{10}}{36{w}^{-4}} & \ \text{Simplify}. \\ & = & \frac{27{w}^{10-(-4)}}{4} & \ \text{The quotient rule and reduce fraction} \\ & = & \frac{27{w}^{14}}{4} & \ \text{Simplify}.\end{array}$

Condensed — the full section is in OpenStax College Algebra 2e.

Using Scientific Notation

Recall at the beginning of the section that we found the number $1.3\times {10}^{13}$ when describing bits of information in digital images. Other extreme numbers include the width of a human hair, which is about 0.00005 m, and the radius of an electron, which is about 0.00000000000047 m. How can we effectively work, read, compare, and calculate with numbers such as these?

A shorthand method of writing very small and very large numbers is called scientific notation, in which we express numbers in terms of exponents of 10. To write a number in scientific notation, move the decimal point to the right of the first digit in the number. Write the digits as a decimal number between 1 and 10. Count the number of places n that you moved the decimal point. Multiply the decimal number by 10 raised to a power of n. If you moved the decimal left as in a very large number, $n$ is positive. If you moved the decimal right as in a very small number, $n$ is negative.

For example, consider the number 2,780,418. Move the decimal left until it is to the right of the first nonzero digit, which is 2.

We obtain 2.780418 by moving the decimal point 6 places to the left. Therefore, the exponent of 10 is 6, and it is positive because we moved the decimal point to the left. This is what we should expect for a large number.

\[2.780418\times {10}^{6}\]

Working with small numbers is similar. Take, for example, the radius of an electron, 0.00000000000047 m. Perform the same series of steps as above, except move the decimal point to the right.

Be careful not to include the leading 0 in your count. We move the decimal point 13 places to the right, so the exponent of 10 is 13. The exponent is negative because we moved the decimal point to the right. This is what we should expect for a small number.

\[4.7\times {10}^{-13}\]

Condensed — the full section is in OpenStax College Algebra 2e.

Key Equations

Rules of Exponents For nonzero real numbers $a$ and $b$ and integers $m$ and $n$
Product rule	${a}^{m}⋅{a}^{n}={a}^{m+n}$
Quotient rule	$\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$
Power rule	${({a}^{m})}^{n}={a}^{m⋅n}$
Zero exponent rule	${a}^{0}=1$
Negative rule	${a}^{-n}=\frac{1}{{a}^{n}}$
Power of a product rule	${(a⋅b)}^{n}={a}^{n}⋅{b}^{n}$
Power of a quotient rule	${(\frac{a}{b})}^{n}=\frac{{a}^{n}}{{b}^{n}}$

Key Concepts

Products of exponential expressions with the same base can be simplified by adding exponents. See .
Quotients of exponential expressions with the same base can be simplified by subtracting exponents. See .
Powers of exponential expressions with the same base can be simplified by multiplying exponents. See .
An expression with exponent zero is defined as 1. See .
An expression with a negative exponent is defined as a reciprocal. See and .
The power of a product of factors is the same as the product of the powers of the same factors. See .
The power of a quotient of factors is the same as the quotient of the powers of the same factors. See .
The rules for exponential expressions can be combined to simplify more complicated expressions. See .
Scientific notation uses powers of 10 to simplify very large or very small numbers. See and .
Scientific notation may be used to simplify calculations with very large or very small numbers. See and .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Write each of the following products with a single base. Do not simplify further.
1. ⓐ ${t}^{5}⋅{t}^{3}$
2. ⓑ ${(-3)}^{5}⋅(-3)$
3. ⓒ ${x}^{2}⋅{x}^{5}⋅{x}^{3}$
Otkrij odgovor
Use the product rule to simplify each expression.
1. ⓐ ${t}^{5}⋅{t}^{3}={t}^{5+3}={t}^{8}$
2. ⓑ ${(-3)}^{5}⋅(-3)={(-3)}^{5}⋅{(-3)}^{1}={(-3)}^{5+1}={(-3)}^{6}$
3. ⓒ ${x}^{2}⋅{x}^{5}⋅{x}^{3}$
At first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two.
\[{x}^{2}⋅{x}^{5}⋅{x}^{3}=({x}^{2}⋅{x}^{5})⋅{x}^{3}=({x}^{2+5})⋅{x}^{3}={x}^{7}⋅{x}^{3}={x}^{7+3}={x}^{10}\]
Notice we get the same result by adding the three exponents in one step.
\[{x}^{2}⋅{x}^{5}⋅{x}^{3}={x}^{2+5+3}={x}^{10}\]
Write each of the following products with a single base. Do not simplify further.
1. ⓐ ${k}^{6}⋅{k}^{9}$
2. ⓑ ${(\frac{2}{y})}^{4}⋅(\frac{2}{y})$
3. ⓒ ${t}^{3}⋅{t}^{6}⋅{t}^{5}$
Otkrij odgovor
1. ⓐ ${k}^{15}$
2. ⓑ ${(\frac{2}{y})}^{5}$
3. ⓒ ${t}^{14}$
Write each of the following products with a single base. Do not simplify further.
1. ⓐ $\frac{{(-2)}^{14}}{{(-2)}^{9}}$
2. ⓑ $\frac{{t}^{23}}{{t}^{15}}$
3. ⓒ $\frac{{(z\sqrt{2})}^{5}}{z\sqrt{2}}$
Otkrij odgovor
Use the quotient rule to simplify each expression.
1. ⓐ $\frac{{(-2)}^{14}}{{(-2)}^{9}}={(-2)}^{14-9}={(-2)}^{5}$
2. ⓑ $\frac{{t}^{23}}{{t}^{15}}={t}^{23-15}={t}^{8}$
3. ⓒ $\frac{{(z\sqrt{2})}^{5}}{z\sqrt{2}}={(z\sqrt{2})}^{5-1}={(z\sqrt{2})}^{4}$
Write each of the following products with a single base. Do not simplify further.
1. ⓐ $\frac{{s}^{75}}{{s}^{68}}$
2. ⓑ $\frac{{(-3)}^{6}}{-3}$
3. ⓒ $\frac{{(e{f}^{2})}^{5}}{{(e{f}^{2})}^{3}}$
Otkrij odgovor
1. ⓐ ${s}^{7}$
2. ⓑ ${(-3)}^{5}$
3. ⓒ ${(e{f}^{2})}^{2}$
Write each of the following products with a single base. Do not simplify further.
1. ⓐ ${({x}^{2})}^{7}$
2. ⓑ ${({(2t)}^{5})}^{3}$
3. ⓒ ${({(-3)}^{5})}^{11}$
Otkrij odgovor
Use the power rule to simplify each expression.
1. ⓐ ${({x}^{2})}^{7}={x}^{2⋅7}={x}^{14}$
2. ⓑ ${({(2t)}^{5})}^{3}={(2t)}^{5⋅3}={(2t)}^{15}$
3. ⓒ ${({(-3)}^{5})}^{11}={(-3)}^{5⋅11}={(-3)}^{55}$
Write each of the following products with a single base. Do not simplify further.
1. ⓐ ${({(3y)}^{8})}^{3}$
2. ⓑ ${({t}^{5})}^{7}$
3. ⓒ ${({(-g)}^{4})}^{4}$
Otkrij odgovor
1. ⓐ ${(3y)}^{24}$
2. ⓑ ${t}^{35}$
3. ⓒ ${(-g)}^{16}$
Simplify each expression using the zero exponent rule of exponents.
1. ⓐ $\frac{{c}^{3}}{{c}^{3}}$
2. ⓑ $\frac{-3{x}^{5}}{{x}^{5}}$
3. ⓒ $\frac{{({j}^{2}k)}^{4}}{({j}^{2}k)⋅{({j}^{2}k)}^{3}}$
4. ⓓ $\frac{5{(r{s}^{2})}^{2}}{{(r{s}^{2})}^{2}}$
Otkrij odgovor

Use the zero exponent and other rules to simplify each expression.
ⓐ
$\begin{array}{lll}\frac{{c}^{3}}{{c}^{3}} & = & {c}^{3-3} \\ & = & {c}^{0} \\ & = & 1\end{array}$
ⓑ
$\begin{array}{lll}\frac{-3{x}^{5}}{{x}^{5}} & = & -3⋅\frac{{x}^{5}}{{x}^{5}} \\ & = & -3⋅{x}^{5-5} \\ & = & -3⋅{x}^{0} \\ & = & -3⋅1 \\ & = & -3\end{array}$
ⓒ
$\begin{array}{llll}\frac{{({j}^{2}k)}^{4}}{({j}^{2}k)⋅{({j}^{2}k)}^{3}} & = & \frac{{({j}^{2}k)}^{4}}{{({j}^{2}k)}^{1+3}} & \ \text{Use the product rule in the denominator}. \\ & = & \frac{{({j}^{2}k)}^{4}}{{({j}^{2}k)}^{4}} & \ \text{Simplify}. \\ & = & {({j}^{2}k)}^{4-4} & \ \text{Use the quotient rule}. \\ & = & {({j}^{2}k)}^{0} & \ \text{Simplify}. \\ & = & 1 & \end{array}$
ⓓ
$\begin{array}{llll}\frac{5{(r{s}^{2})}^{2}}{{(r{s}^{2})}^{2}} & = & 5{(r{s}^{2})}^{2-2} & \ \text{Use the quotient rule}. \\ & = & 5{(r{s}^{2})}^{0} & \ \text{Simplify}. \\ & = & 5⋅1 & \ \text{Use the zero exponent rule}. \\ & = & 5 & \ \text{Simplify}.\end{array}$
Simplify each expression using the zero exponent rule of exponents.
1. ⓐ $\frac{{t}^{7}}{{t}^{7}}$
2. ⓑ $\frac{{(d{e}^{2})}^{11}}{2{(d{e}^{2})}^{11}}$
3. ⓒ $\frac{{w}^{4}⋅{w}^{2}}{{w}^{6}}$
4. ⓓ $\frac{{t}^{3}⋅{t}^{4}}{{t}^{2}⋅{t}^{5}}$
Otkrij odgovor
1. ⓐ $1$
2. ⓑ $\frac{1}{2}$
3. ⓒ $1$
4. ⓓ $1$
Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
1. ⓐ $\frac{{\theta }^{3}}{{\theta }^{10}}$
2. ⓑ $\frac{{z}^{2}⋅z}{{z}^{4}}$
3. ⓒ $\frac{{(-5{t}^{3})}^{4}}{{(-5{t}^{3})}^{8}}$
Otkrij odgovor
1. ⓐ $\frac{{\theta }^{3}}{{\theta }^{10}}={\theta }^{3-10}={\theta }^{-7}=\frac{1}{{\theta }^{7}}$
2. ⓑ $\frac{{z}^{2}⋅z}{{z}^{4}}=\frac{{z}^{2+1}}{{z}^{4}}=\frac{{z}^{3}}{{z}^{4}}={z}^{3-4}={z}^{-1}=\frac{1}{z}$
3. ⓒ $\frac{{(-5{t}^{3})}^{4}}{{(-5{t}^{3})}^{8}}={(-5{t}^{3})}^{4-8}={(-5{t}^{3})}^{-4}=\frac{1}{{(-5{t}^{3})}^{4}}$
Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
1. ⓐ $\frac{{(-3t)}^{2}}{{(-3t)}^{8}}$
2. ⓑ $\frac{{f}^{47}}{{f}^{49}⋅f}$
3. ⓒ $\frac{2{k}^{4}}{5{k}^{7}}$
Otkrij odgovor
1. ⓐ $\frac{1}{{(-3t)}^{6}}$
2. ⓑ $\frac{1}{{f}^{3}}$
3. ⓒ $\frac{2}{5{k}^{3}}$
Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
1. ⓐ ${b}^{2}⋅{b}^{-8}$
2. ⓑ ${(-x)}^{5}⋅{(-x)}^{-5}$
3. ⓒ $\frac{-7z}{{(-7z)}^{5}}$
Otkrij odgovor
1. ⓐ ${b}^{2}⋅{b}^{-8}={b}^{2-8}={b}^{-6}=\frac{1}{{b}^{6}}$
2. ⓑ ${(-x)}^{5}⋅{(-x)}^{-5}={(-x)}^{5-5}={(-x)}^{0}=1$
3. ⓒ $\frac{-7z}{{(-7z)}^{5}}=\frac{{(-7z)}^{1}}{{(-7z)}^{5}}={(-7z)}^{1-5}={(-7z)}^{-4}=\frac{1}{{(-7z)}^{4}}$
Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents.
1. ⓐ ${t}^{-11}⋅{t}^{6}$
2. ⓑ $\frac{{25}^{12}}{{25}^{13}}$
Otkrij odgovor
1. ⓐ ${t}^{-5}=\frac{1}{{t}^{5}}$
2. ⓑ $\frac{1}{25}$
Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
1. ⓐ ${(a{b}^{2})}^{3}$
2. ⓑ ${(2t)}^{15}$
3. ⓒ ${(-2{w}^{3})}^{3}$
4. ⓓ $\frac{1}{{(-7z)}^{4}}$
5. ⓔ ${({e}^{-2}{f}^{2})}^{7}$
Otkrij odgovor
Use the product and quotient rules and the new definitions to simplify each expression.
1. ⓐ ${(a{b}^{2})}^{3}={(a)}^{3}⋅{({b}^{2})}^{3}={a}^{1⋅3}⋅{b}^{2⋅3}={a}^{3}{b}^{6}$
2. ⓑ ${(2t)}^{15}={(2)}^{15}⋅{(t)}^{15}={2}^{15}{t}^{15}=32,768{t}^{15}$
3. ⓒ ${(-2{w}^{3})}^{3}={(-2)}^{3}⋅{({w}^{3})}^{3}=-8⋅{w}^{3⋅3}=-8{w}^{9}$
4. ⓓ $\frac{1}{{(-7z)}^{4}}=\frac{1}{{(-7)}^{4}⋅{(z)}^{4}}=\frac{1}{2,401{z}^{4}}$
5. ⓔ ${({e}^{-2}{f}^{2})}^{7}={({e}^{-2})}^{7}⋅{({f}^{2})}^{7}={e}^{-2⋅7}⋅{f}^{2⋅7}={e}^{-14}{f}^{14}=\frac{{f}^{14}}{{e}^{14}}$
Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents.
1. ⓐ ${({g}^{2}{h}^{3})}^{5}$
2. ⓑ ${(5t)}^{3}$
3. ⓒ ${(-3{y}^{5})}^{3}$
4. ⓓ $\frac{1}{{({a}^{6}{b}^{7})}^{3}}$
5. ⓔ ${({r}^{3}{s}^{-2})}^{4}$
Otkrij odgovor
1. ⓐ ${g}^{10}{h}^{15}$
2. ⓑ $125{t}^{3}$
3. ⓒ $-27{y}^{15}$
4. ⓓ $\frac{1}{{a}^{18}{b}^{21}}$
5. ⓔ $\frac{{r}^{12}}{{s}^{8}}$
Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.
1. ⓐ ${(\frac{4}{{z}^{11}})}^{3}$
2. ⓑ ${(\frac{p}{{q}^{3}})}^{6}$
3. ⓒ ${(\frac{-1}{{t}^{2}})}^{27}$
4. ⓓ ${({j}^{3}{k}^{-2})}^{4}$
5. ⓔ ${({m}^{-2}{n}^{-2})}^{3}$
Otkrij odgovor
1. ⓐ ${(\frac{4}{{z}^{11}})}^{3}=\frac{{(4)}^{3}}{{({z}^{11})}^{3}}=\frac{64}{{z}^{11⋅3}}=\frac{64}{{z}^{33}}$
2. ⓑ ${(\frac{p}{{q}^{3}})}^{6}=\frac{{(p)}^{6}}{{({q}^{3})}^{6}}=\frac{{p}^{1⋅6}}{{q}^{3⋅6}}=\frac{{p}^{6}}{{q}^{18}}$
3. ⓒ ${(\frac{-1}{{t}^{2}})}^{27}=\frac{{(-1)}^{27}}{{({t}^{2})}^{27}}=\frac{-1}{{t}^{2⋅27}}=\frac{-1}{{t}^{54}}=-\frac{1}{{t}^{54}}$
4. ⓓ ${({j}^{3}{k}^{-2})}^{4}={(\frac{{j}^{3}}{{k}^{2}})}^{4}=\frac{{({j}^{3})}^{4}}{{({k}^{2})}^{4}}=\frac{{j}^{3⋅4}}{{k}^{2⋅4}}=\frac{{j}^{12}}{{k}^{8}}$
5. ⓔ ${({m}^{-2}{n}^{-2})}^{3}={(\frac{1}{{m}^{2}{n}^{2}})}^{3}=\frac{{(1)}^{3}}{{({m}^{2}{n}^{2})}^{3}}=\frac{1}{{({m}^{2})}^{3}{({n}^{2})}^{3}}=\frac{1}{{m}^{2⋅3}⋅{n}^{2⋅3}}=\frac{1}{{m}^{6}{n}^{6}}$
Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents.
1. ⓐ ${(\frac{{b}^{5}}{c})}^{3}$
2. ⓑ ${(\frac{5}{{u}^{8}})}^{4}$
3. ⓒ ${(\frac{-1}{{w}^{3}})}^{35}$
4. ⓓ ${({p}^{-4}{q}^{3})}^{8}$
5. ⓔ ${({c}^{-5}{d}^{-3})}^{4}$
Otkrij odgovor
1. ⓐ $\frac{{b}^{15}}{{c}^{3}}$
2. ⓑ $\frac{625}{{u}^{32}}$
3. ⓒ $\frac{-1}{{w}^{105}}$
4. ⓓ $\frac{{q}^{24}}{{p}^{32}}$
5. ⓔ $\frac{1}{{c}^{20}{d}^{12}}$
Simplify each expression and write the answer with positive exponents only.
1. ⓐ ${(6{m}^{2}{n}^{-1})}^{3}$
2. ⓑ ${17}^{5}⋅{17}^{-4}⋅{17}^{-3}$
3. ⓒ ${(\frac{{u}^{-1}v}{{v}^{-1}})}^{2}$
4. ⓓ $(-2{a}^{3}{b}^{-1})(5{a}^{-2}{b}^{2})$
5. ⓔ ${({x}^{2}\sqrt{2})}^{4}{({x}^{2}\sqrt{2})}^{-4}$
6. ⓕ $\frac{{(3{w}^{2})}^{5}}{{(6{w}^{-2})}^{2}}$
Otkrij odgovor
1. ⓐ
  $\begin{array}{llll}{(6{m}^{2}{n}^{-1})}^{3} & = & {(6)}^{3}{({m}^{2})}^{3}{({n}^{-1})}^{3} & \ \text{The power of a product rule} \\ & = & {6}^{3}{m}^{2⋅3}{n}^{-1⋅3} & \ \text{The power rule} \\ & = & \ 216{m}^{6}{n}^{-3} & \ \text{Simplify}. \\ & = & \frac{216{m}^{6}}{{n}^{3}} & \ \text{The negative exponent rule}\end{array}$
2. ⓑ
  $\begin{array}{llll}{17}^{5}⋅{17}^{-4}⋅{17}^{-3} & = & {17}^{5-4-3} & \ \text{The product rule} \\ & = & {17}^{-2} & \ \text{Simplify}. \\ & = & \frac{1}{{17}^{2}}\ \text{or }\frac{1}{289} & \ \text{The negative exponent rule}\end{array}$
3. ⓒ
  $\begin{array}{llll}{(\frac{{u}^{-1}v}{{v}^{-1}})}^{2} & = & \frac{{({u}^{-1}v)}^{2}}{{({v}^{-1})}^{2}} & \ \text{The power of a quotient rule} \\ & = & \frac{{u}^{-2}{v}^{2}}{{v}^{-2}} & \ \text{The power of a product rule} \\ & = & {u}^{-2}{v}^{2-(-2)} & \ \text{The quotient rule} \\ & = & {u}^{-2}{v}^{4} & \ \text{Simplify}. \\ & = & \frac{{v}^{4}}{{u}^{2}} & \ \text{The negative exponent rule}\end{array}$
4. ⓓ
  $\begin{array}{llll}(-2{a}^{3}{b}^{-1})(5{a}^{-2}{b}^{2}) & = & -2⋅5⋅{a}^{3}⋅{a}^{-2}⋅{b}^{-1}⋅{b}^{2} & \ \text{Commutative and associative laws of multiplication} \\ & = & -10⋅{a}^{3-2}⋅{b}^{-1+2} & \ \text{The product rule} \\ & = & -10ab & \ \text{Simplify}.\end{array}$
5. ⓔ
  $\begin{array}{llll}{({x}^{2}\sqrt{2})}^{4}{({x}^{2}\sqrt{2})}^{-4} & = & {({x}^{2}\sqrt{2})}^{4-4} & \ \text{The product rule} \\ & = & \ {({x}^{2}\sqrt{2})}^{0} & \ \text{Simplify}. \\ & = & 1 & \ \text{The zero exponent rule}\end{array}$
6. ⓕ
  $\begin{array}{llll}\frac{{(3{w}^{2})}^{5}}{{(6{w}^{-2})}^{2}} & = & \frac{{(3)}^{5}⋅{({w}^{2})}^{5}}{{(6)}^{2}⋅{({w}^{-2})}^{2}} & \ \text{The power of a product rule} \\ & = & \frac{{3}^{5}{w}^{2⋅5}}{{6}^{2}{w}^{-2⋅2}} & \ \text{The power rule} \\ & = & \frac{243{w}^{10}}{36{w}^{-4}} & \ \text{Simplify}. \\ & = & \frac{27{w}^{10-(-4)}}{4} & \ \text{The quotient rule and reduce fraction} \\ & = & \frac{27{w}^{14}}{4} & \ \text{Simplify}.\end{array}$
Simplify each expression and write the answer with positive exponents only.
1. ⓐ ${(2u{v}^{-2})}^{-3}$
2. ⓑ ${x}^{8}⋅{x}^{-12}⋅x$
3. ⓒ ${(\frac{{e}^{2}{f}^{-3}}{{f}^{-1}})}^{2}$
4. ⓓ $(9{r}^{-5}{s}^{3})(3{r}^{6}{s}^{-4})$
5. ⓔ ${(\frac{4}{9}t{w}^{-2})}^{-3}{(\frac{4}{9}t{w}^{-2})}^{3}$
6. ⓕ $\frac{{(2{h}^{2}k)}^{4}}{{(7{h}^{-1}{k}^{2})}^{2}}$
Otkrij odgovor
1. ⓐ $\frac{{v}^{6}}{8{u}^{3}}$
2. ⓑ $\frac{1}{{x}^{3}}$
3. ⓒ $\frac{{e}^{4}}{{f}^{4}}$
4. ⓓ $\frac{27r}{s}$
5. ⓔ $1$
6. ⓕ $\frac{16{h}^{10}}{49}$
Write each number in scientific notation.
ⓐDistance to Andromeda Galaxy from Earth: 24,000,000,000,000,000,000,000 m
ⓑDiameter of Andromeda Galaxy: 1,300,000,000,000,000,000,000 m
ⓒNumber of stars in Andromeda Galaxy: 1,000,000,000,000
ⓓDiameter of electron: 0.00000000000094 m
ⓔProbability of being struck by lightning in any single year: 0.00000143

Otkrij odgovor

ⓐ
$\begin{array}{l}24,000,000,000,000,000,000,000\ \text{m} \\ \underset{←22\ \text{places}}{\underset{}{24,000,000,000,000,000,000,000\ \text{m}}} \\ 2.4\times {10}^{22}\ \text{m}\end{array}$
ⓑ
$\begin{array}{l}1,300,000,000,000,000,000,000\ \text{m} \\ \underset{←21\ \text{places}}{\underset{}{1,300,000,000,000,000,000,000\ \text{m}}} \\ 1.3\times {10}^{21}\ \text{m}\end{array}$
ⓒ
$\begin{array}{l}1,000,000,000,000 \\ \underset{←12\ \text{places}}{\underset{}{1,000,000,000,000}} \\ 1\times {10}^{12}\end{array}$
ⓓ
$\begin{array}{l}0.00000000000094\ \text{m} \\ \underset{\to 13\ \text{places}}{\underset{}{0.00000000000094\ \text{m}}} \\ 9.4\times {10}^{-13}\ \text{m}\end{array}$
ⓔ
$\begin{array}{l}0.00000143 \\ \underset{\to 6\ \text{places}}{\underset{}{0.00000143}} \\ 1.43\times {10}^{-6}\end{array}$
Write each number in scientific notation.
1. ⓐU.S. national debt per taxpayer (April 2014): $152,000
2. ⓑWorld population (April 2014): 7,158,000,000
3. ⓒWorld gross national income (April 2014): $85,500,000,000,000
4. ⓓTime for light to travel 1 m: 0.00000000334 s
5. ⓔProbability of winning lottery (match 6 of 49 possible numbers): 0.0000000715
Otkrij odgovor
1. ⓐ $\$1.52\times {10}^{5}$
2. ⓑ $7.158\times {10}^{9}$
3. ⓒ $\$8.55\times {10}^{13}$
4. ⓓ $3.34\times {10}^{-9}$
5. ⓔ $7.15\times {10}^{-8}$
Convert each number in scientific notation to standard notation.
1. ⓐ $3.547\times {10}^{14}$
2. ⓑ $-2\times {10}^{6}$
3. ⓒ $7.91\times {10}^{-7}$
4. ⓓ $-8.05\times {10}^{-12}$
Otkrij odgovor
1. ⓐ
  $\begin{array}{l}3.547\times {10}^{14} \\ \underset{\to 14\ \text{places}}{\underset{}{3.54700000000000}} \\ 354,700,000,000,000\end{array}$
2. ⓑ
  $\begin{array}{l}-2\times {10}^{6} \\ \underset{\to 6\ \text{places}}{\underset{}{-2.000000}} \\ -2,000,000\end{array}$
3. ⓒ
  $\begin{array}{l}7.91\times {10}^{-7} \\ \underset{←7\ \text{places}}{\underset{}{0000007.91}} \\ 0.000000791\end{array}$
4. ⓓ
  $\begin{array}{l}-8.05\times {10}^{-12} \\ \underset{←12\ \text{places}}{\underset{}{-000000000008.05}} \\ -0.00000000000805\end{array}$
Convert each number in scientific notation to standard notation.
1. ⓐ $7.03\times {10}^{5}$
2. ⓑ $-8.16\times {10}^{11}$
3. ⓒ $-3.9\times {10}^{-13}$
4. ⓓ $8\times {10}^{-6}$
Otkrij odgovor
1. ⓐ $703,000$
2. ⓑ $-816,000,000,000$
3. ⓒ $-0.000\ 000\ 000\ 000\ 39$
4. ⓓ $0.000008$
Perform the operations and write the answer in scientific notation.
1. ⓐ $(8.14\times {10}^{-7})(6.5\times {10}^{10})$
2. ⓑ $(4\times {10}^{5})\div (-1.52\times {10}^{9})$
3. ⓒ $(2.7\times {10}^{5})(6.04\times {10}^{13})$
4. ⓓ $(1.2\times {10}^{8})\div (9.6\times {10}^{5})$
5. ⓔ $(3.33\times {10}^{4})(-1.05\times {10}^{7})(5.62\times {10}^{5})$
Otkrij odgovor
1. ⓐ
  $\begin{array}{llll}(8.14\times {10}^{-7})(6.5\times {10}^{10}) & = & (8.14\times 6.5)({10}^{-7}\times {10}^{10}) & \begin{array}{l}\ \text{Commutative and associative} \\ \text{properties of multiplication}\end{array} \\ & = & (52.91)({10}^{3}) & \ \text{Product rule of exponents} \\ & = & 5.291\times {10}^{4} & \ \text{Scientific notation}\end{array}$
2. ⓑ
  $\begin{array}{llll}(4\times {10}^{5})\div (-1.52\times {10}^{9}) & = & (\frac{4}{-1.52})(\frac{{10}^{5}}{{10}^{9}}) & \begin{array}{l}\ \text{Commutative and associative} \\ \text{properties of multiplication}\end{array} \\ & \approx & (-2.63)({10}^{-4}) & \ \text{Quotient rule of exponents} \\ & = & -2.63\times {10}^{-4} & \ \text{Scientific notation}\end{array}$
3. ⓒ
  $\begin{array}{llll}(2.7\times {10}^{5})(6.04\times {10}^{13}) & = & (2.7\times 6.04)({10}^{5}\times {10}^{13}) & \begin{array}{l}\ \text{Commutative and associative} \\ \text{properties of multiplication}\end{array} \\ & = & (16.308)({10}^{18}) & \ \text{Product rule of exponents} \\ & = & 1.6308\times {10}^{19} & \ \text{Scientific notation}\end{array}$
4. ⓓ
  $\begin{array}{llll}(1.2\times {10}^{8})\div (9.6\times {10}^{5}) & = & (\frac{1.2}{9.6})(\frac{{10}^{8}}{{10}^{5}}) & \begin{array}{l}\ \text{Commutative and associative} \\ \text{properties of multiplication}\end{array} \\ & = & (0.125)({10}^{3}) & \ \text{Quotient rule of exponents} \\ & = & 1.25\times {10}^{2} & \ \text{Scientific notation}\end{array}$
5. ⓔ
  $\begin{array}{lll}(3.33\times {10}^{4})(-1.05\times {10}^{7})(5.62\times {10}^{5}) & = & [3.33\times (-1.05)\times 5.62]({10}^{4}\times {10}^{7}\times {10}^{5}) \\ & \approx & (-19.65)({10}^{16}) \\ & = & -1.965\times {10}^{17}\end{array}$
Perform the operations and write the answer in scientific notation.
1. ⓐ $(-7.5\times {10}^{8})(1.13\times {10}^{-2})$
2. ⓑ $(1.24\times {10}^{11})\div (1.55\times {10}^{18})$
3. ⓒ $(3.72\times {10}^{9})(8\times {10}^{3})$
4. ⓓ $(9.933\times {10}^{23})\div (-2.31\times {10}^{17})$
5. ⓔ $(-6.04\times {10}^{9})(7.3\times {10}^{2})(-2.81\times {10}^{2})$
Otkrij odgovor
1. ⓐ $-8.475\times {10}^{6}$
2. ⓑ $8\times {10}^{-8}$
3. ⓒ $2.976\times {10}^{13}$
4. ⓓ $-4.3\times {10}^{6}$
5. ⓔ $\approx 1.24\times {10}^{15}$
In April 2014, the population of the United States was about 308,000,000 people. The national debt was about $17,547,000,000,000. Write each number in scientific notation, rounding figures to two decimal places, and find the amount of the debt per U.S. citizen. Write the answer in both scientific and standard notations.

Otkrij odgovor

The population was $308,000,000=3.08\times {10}^{8}.$
The national debt was $\text{\$}17,547,000,000,000\approx \text{\$}1.75\times {10}^{13}.$
To find the amount of debt per citizen, divide the national debt by the number of citizens.
\[\begin{array}{lll}(1.75\times {10}^{13})\div (3.08\times {10}^{8}) & = & (\frac{1.75}{3.08})⋅(\frac{{10}^{13}}{{10}^{8}}) \\ & \approx & 0.57\times {10}^{5} \\ & = & 5.7\times {10}^{4}\end{array}\]
The debt per citizen at the time was about $\text{\$}5.7\times {10}^{4},$ or $57,000.
An average human body contains around 30,000,000,000,000 red blood cells. Each cell measures approximately 0.000008 m long. Write each number in scientific notation and find the total length if the cells were laid end-to-end. Write the answer in both scientific and standard notations.

Otkrij odgovor

Number of cells: $3\times {10}^{13};$ length of a cell: $8\times {10}^{-6}$ m; total length: $2.4\times {10}^{8}$ m or $240,000,000$ m.
Is ${2}^{3}$ the same as ${3}^{2}?$ Explain.

Otkrij odgovor

No, the two expressions are not the same. An exponent tells how many times you multiply the base. So ${2}^{3}$ is the same as $2\times 2\times 2,$ which is 8. ${3}^{2}$ is the same as $3\times 3,$ which is 9.
When can you add two exponents?
What is the purpose of scientific notation?

Otkrij odgovor

It is a method of writing very small and very large numbers.
Explain what a negative exponent does.
${9}^{2}$

Otkrij odgovor

81
${15}^{-2}$
${3}^{2}\times {3}^{3}$

Otkrij odgovor

243
${4}^{4}\div 4$
${({2}^{2})}^{-2}$

Otkrij odgovor

$\frac{1}{16}$
${(5-8)}^{0}$
${11}^{3}\div {11}^{4}$

Otkrij odgovor

$\frac{1}{11}$
${6}^{5}\times {6}^{-7}$
${({8}^{0})}^{2}$

Otkrij odgovor

1
${5}^{-2}\div {5}^{2}$

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\theta

theta
The usual name for an angle.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Exponents and Scientific Notation

Use the product rule of exponents.
Use the quotient rule of exponents.
Use the power rule of exponents.
Use the zero exponent rule of exponents.
Use the negative rule of exponents.
Find the power of a product and a quotient.
Simplify exponential expressions.
Use scientific notation.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Više u Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value

Rules of Exponents For nonzero real numbers \(a\) and \(b\) and integers \(m\) and \(n\)
Product rule	\({a}^{m}⋅{a}^{n}={a}^{m+n}\)
Quotient rule	\(\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}\)
Power rule	\({({a}^{m})}^{n}={a}^{m⋅n}\)
Zero exponent rule	\({a}^{0}=1\)
Negative rule	\({a}^{-n}=\frac{1}{{a}^{n}}\)
Power of a product rule	\({(a⋅b)}^{n}={a}^{n}⋅{b}^{n}\)
Power of a quotient rule	\({(\frac{a}{b})}^{n}=\frac{{a}^{n}}{{b}^{n}}\)