1.1 Exponents and Scientific Notation

Terri Manthey

1.1 Exponents and Scientific Notation

Review of Intermediate Algebra

Mathematicians, scientists, economists and other professionals commonly encounter very large and very small numbers. However, it may not be obvious how common values like these are in everyday life. For instance, a pixel is the smallest unit of light that can be perceived and recorded by a digital camera. A particular camera might record an image that is 2,048 pixels by 1,536 pixels, which is a very high-resolution picture. It can also perceive a color depth (gradations in colors) of up to 48 bits per frame and can shoot the equivalent of 24 frames per second. The maximum possible number of bits of information used to film a one-hour (3,600-second) digital film is then an extremely large number.

Using a calculator, we enter 2,048 × 1,536 × 48 × 24 × 3,600 and press ENTER. The calculator displays 1.304596316E13. What does this mean? The “E13” portion of the result represents the exponent 13 of ten, so there are a maximum of approximately 1.3×10¹³ bits of data in that one-hour film. In this section, we review rules of exponents first and then apply them to calculations involving very large or small numbers.

Using the Product Rule of Exponents

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

$\begin{array}{l} \mathop {\left( {x \cdot x \cdot x} \right)}\limits^{\text{3 \emph{factors}}} \cdot \mathop {\left( {x \cdot x \cdot x \cdot x} \right)}\limits^{\text{4 \emph{factors}}}\\ \quad \quad \;\; = \mathop {x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x}\limits^{\text{7 \emph{factors}}}\\ \quad \quad \;\; = {x^7} \end{array}$

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

Notice that the exponent of the product is the sum of the exponents of the factors. 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.

Product Rule of Exponents

For any real number a and natural numbers m and n, the product rule of exponents states that:

${a^m} \cdot {a^n} = {a^{m + n}}$

Now consider an example with real numbers.

${2^3} \cdot {2^4} = {2^{3 + 4}} = {2^7}$

We can always check that this is true by simplifying each exponential expression. We find that ${2^3} = 8,\;{2^4} = 16$ and ${2^7} = 128.$ The product $8 \cdot 16 = 128$ 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. We can even use the product rule of exponents with integers as well, as seen in example c below.

Examples Using the Product Rule of Exponents

a. ${t^5} \cdot {t^3}$

Here, we would just add the exponents:
${t^{5 + 3}} = {t^8}$

b. ${( - 3)^5}\cdot {( - 3)}$

This seems complicated because of the negative, but again, once we add the exponents, this would become positive due to the even exponent. (Note that when we do not see an exponent, we assume that it is 1.)

${( - 3)^{5 + 1}} = {( - 3)^6} = 729$

c. ${x^2} \cdot {x^5} \cdot {x^{-3}}$

At first, we may think we cannot simplify a product of three factors, but using the associative properties of multiplication and addition, we can simply add all three exponents. Since the 3 is negative, we add a negative 3, which is the same as subtracting 3.

${x^{2 + 5 - 3}} = {x^{4}}$

Try it Now 1

a. ${k^6} \cdot {k^9}$

b. $\left( {\dfrac{2}{y}} \right)^{4}} \cdot \left( {\dfrac{2}{y}} \right)$

c. ${t^{ - 3}} \cdot {t^6} \cdot {t^5}$

See Try it Now answers at the end of each section.

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. In a similar way to the product rule, we can simplify an expression such as $\dfrac{y^m}{y^n}$ . Consider the example $\dfrac {y^6}{y^2}$ . What happens if we write this out:

$\begin{array}{l} \dfrac{{{y^6}}}{{{y^2}}} = \dfrac{{y \cdot y \cdot y \cdot y \cdot y \cdot y}}{{y \cdot y}}\\ \quad \; = \dfrac{{ \cancel{y}\cdot \cancel{y} \cdot y \cdot y \cdot y \cdot y}}{{ \cancel{y}\cdot \cancel{y}}}\\ \quad \; = \dfrac{{y \cdot y \cdot y \cdot y}}{1} = {y^4} \end{array}$

When dividing exponential expressions with the same base, we write the result with the common base and subtract the exponents.

$\dfrac{y^6}{y^2}=y^{6-2}=y^4$

Quotient Rule for Exponents

For any nonzero real number $a$ and natural numbers $m$ and $n$ , the quotient rule of exponents states that

$\dfrac{a^m}{a^n}=a^{m-n}$

Examples using the Quotient Rule of Exponents

a. $\dfrac{(-2)^7}{(-2)^4}$

Here, we can just subtract the exponents: $(-2)^{7-4}=(-2)^{3}=-8$

b. $\dfrac{t^{28}}{t^{16}}$

This might look difficult, but again, a simple subtraction: $t^{28-16}=t^{12}$

c. $\dfrac{\left(z\sqrt{2} \right)^5}{z\sqrt{2}}$

We begin with the quotient rules of exponents: $\left(z\sqrt{2}\right)^{5-1}=\left(z\sqrt{2}\right)^{4}$ (this can be simplified further so we will revisit it later)

Try it Now 2

a. $\dfrac{c^{16}}{c^{9}}$

b. $\dfrac{\left({-3}\right)^{4}}{-3}$

c. $\dfrac{\left({uv}\right)^{5}}{\left({uv} \right) ^{4}}$

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 $\left( x^3 \right)^2$ . If we break up the outside exponent, it means that $x^3$ is multiplied to itself twice, but as we’ve seen above with the Product Rule, we can add the exponents. And note that repeated addition is just multiplication.

${\left( {{x^3}} \right)^2} = {x^3} \cdot {x^3} = {x^{3 + 3}} = {x^{2 \cdot 3}} = {x^6}$

The exponent of the answer is the product of the exponents: $\left( x^3 \right)^2 = x^{3 \cdot 2}=x^6$ 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.

The Power Rule of Exponents

For any real number $a$ and positive integers $m$ and $n$ , the power rule of exponents states that:

${\left( {{a^m}} \right)^n} = {a^{m\cdot n}} = {a^{mn}}$

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.

	Product Rule				Power Rule
$5^3 \cdot 5^4$	$=5^{3+4}$	$=5^7$	but	$\left( 5^3 \right) ^4$	$=5^{3 \cdot 4}$	$=5^{12}$
$x^5 \cdot x^2$	$=x^{5+2}$	$=x^7$	but	$\left( x^5 \right) ^2$	$=x^{5 \cdot 2}$	$=x^{10}$
$\left( 3a \right)^7 \cdot \left( 3a \right) ^{10}$	$=\left(3a \right)^{7+10}$	$=\left(3a \right)^{17}$	but	$\left( \left(3a \right) ^7 \right) ^{10}$	$=\left( 3a \right) ^{7 \cdot 10}$	$=\left(3a \right)^{70}$

Examples using the Power Rule of exponents

a. $\left( x^3 \right)^5$

Just multiply the exponents: $\left( x^3 \right)^5 = x^{3 \cdot 5}=x^{15}$

b. $\left( \left( 2t \right)^2 \right) ^7$

Again, we just multiply the exponents but make sure we keep the parenthesis around 2t so that we remember that the exponent applies to both 2 and t.

$\left( \left( 2t \right)^2 \right) ^7 = \left( 2t \right)^{2 \cdot 7}=\left( 2t \right)^{14}$

c. $\left( \left( -3 \right)^5 \right) ^{10}$

$\left( \left( -3 \right)^5 \right) ^{10}=\left( -3 \right)^{5 \cdot 10} = \left( -3 \right)^{50}$

But, does it really still have to be negative? Remember a negative number to an even power is positive! So we can call this $3^{50}$ which is a really large number!

Try It Now 3

a. $\left( \left( 3y \right)^8 \right) ^3$

b. $\left( t^5 \right)^7$

c. $\left( \left( -g \right)^4 \right) ^{4}$

Using the Zero Exponent Rule of Exponents

Return to the quotient rule. 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.

$\dfrac{t^8}{t^8} = \dfrac {\cancel{t^8}}{\cancel{t^8}}= 1$

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

$\dfrac{t^8}{t^8} = t^{8-8}= t^0$

If we take the two answers equal to each other, we would have $t^0=1$ . This is true for any non-zero real number base or a variable that represents a non-zero real number. The exception is $0^0$ . This is an indeterminate form.

Zero Exponent Rule

For any nonzero real number $a$ , the zero exponent rule of exponents states that:

$a^0= 1$

Examples Using the Zero Exponent Rule

a. $\dfrac{{{c^3}}}{{{c^3}}}$

$c^{3-3}=c^0=1$

b. $\dfrac{{-3x^5}}{{x^5}}$

The -3 here just comes along for the ride: $-3x^{5-5}=-3x^0=-3 \cdot 1=-3$

c. $\dfrac{\left(j^2k \right)^4}{\left(j^2k \right) \cdot \left(j^2k \right)^3}$

This will take two steps, first we can use the addition rule for the denominator:

$\dfrac{\left(j^2k \right)^4}{\left(j^2k \right) \cdot \left(j^2k \right)^3} = \dfrac{\left(j^2k \right) ^4}{\left(j^2k \right) ^{1+3}}=\dfrac{\left(j^2k \right) ^4}{\left(j^2k \right) ^4}$

Now we apply the zero exponent rule:

$\dfrac{\left(j^2k \right) ^4}{\left(j^2k \right) ^4}=\left(j^2k \right) ^{4-4}=\left(j^2k \right) ^0=1$

Try It Now 4

a. $\dfrac{{{x^6}}}{{{x^6}}}$

b. $\dfrac{\left(dn^2 \right) ^{11}}{2\left(dn^2 \right) ^{11}}$

c. $\dfrac{w^4 \cdot w^3}{w^7}$

Using the Negative Rule of Exponents

All our previous examples in the quotient rule had the exponent in the numerator larger than the denominator. In the zero exponent rule, we investigated what happened when the exponents in the numerator and denominator were equal. Another useful result occurs if we consider what happens when $m<n$ . For example, can we simplify $\dfrac{h^3}{h^5}$ ?

Use our example, $\dfrac{h^3}{h^5}$ .

$\begin{array}{l} \dfrac{h^3}{h^5} = \dfrac{h \cdot h \cdot h}{h \cdot h \cdot h \cdot h \cdot h}\\ \quad \; = \dfrac{ \cancel{h} \cdot \cancel{h} \cdot \cancel{h}}{ \cancel{h} \cdot \cancel{h} \cdot \cancel{h} \cdot h \cdot h}\\ \quad \; = \dfrac{1}{h \cdot h} = \dfrac{1}{h^2} \end{array}$

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

$\dfrac{h^3}{h^5} = h^{3-5} =h^{-2}$

Putting the answers together, we have $h^{-2}=\dfrac{1}{h^2}$ . This is true for any nonzero real number, or any variable representing a nonzero real number. Why do we have to say “nonzero?” Because we cannot have 0 in the denominator of a fraction. $\dfrac{1}{0}$ is undefined.

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.

$a^{-n}=\dfrac{1}{a^n} \textrm{ and } a^n=\dfrac{1}{a^{-n}}$

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 $n$ .

The Negative Rule of Exponents

For any nonzero real number $a$ and natural number $n$ the negative rule of exponents states that

$a^{-n}=\dfrac{1}{a^n}$

Examples Using the Quotient Rule with the Negative Exponent Rule

a. $\dfrac {b^3}{b^{10}}$

Just subtract the exponents: $\dfrac {b^3}{b^{10}}=b^{3-10}=b^{-7}$

Now use the negative exponent rule to make this a positive exponent: $b^{-7}=\dfrac{1}{b^7}$

b. $\dfrac{z^2 \cdot z}{z^4}$

We start by combining the two bases in the numerator using the product rule: $\dfrac{z^2 \cdot z}{z^4}=\dfrac{z^3}{z^4}$

Now we use the quotient rule: $\dfrac{z^3}{z^4}=z^{3-4}=z^{-1}$

Now we use negative exponent rule: $z^{-1}=\dfrac{1}{z^1}=\dfrac{1}{z}$

c. $\dfrac{\left( -5t^3 \right)^4}{\left(-5t^3\right)^8}$

Use our quotient rule and negative exponent rule:

$\dfrac{\left( -5t^3 \right)^4}{\left(-5t^3\right)^8}=\left( -5t^3 \right)^{4-8}=\left( -5t^3 \right)^{-4}=\dfrac{1}{\left( -5t^3 \right)^{4}}$

Try it Now 5

a. $\dfrac{x^3}{x^9}$

b. $\dfrac{t^{12}}{t^{15}\cdot t}$

c. $\dfrac{2k^2}{5k^5}$

Examples using the Product and Quotient Rules and the Zero and Negative Exponent Rule

a. $b^2 \cdot b^{-8}$

Start with the product rule: $b^2 \cdot b^{-8}=b^{2+(-8)}=b^{2-8}=b^{-6}$

Now the negative exponent rule: $b^{-6}= \dfrac{1}{b^6}$

b. $\left(-x\right)^{5} \cdot \left(-x\right)^{-5}$

$\left(-x\right)^{5} \cdot \left(-x\right)^{-5}=\left(-x\right)^{5+-5}=\left(-x\right)^{5-5}=\left(-x\right)^{0}=1$

c. $\dfrac{-2z}{\left( -2z \right)^{-3}}$

Perform the quotient rule:

$\dfrac{-2z}{\left( -2z \right)^{-3}}=\left( -2z \right)^{1-(-3)}=\left( -2z \right)^{1+3}=\left( -2z \right)^{4}$

Try it Now 6

a. $t^{-11} \cdot t^{8}$

b. $\dfrac {x^{24}}{x^{25}}$

Find 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 $\left(xy\right)^3$ We begin by using the associative and commutative properties of multiplication to regroup the factors.

$\begin{align*} (xy)^3&=\mathop{(xy)\cdot(xy)\cdot(xy)}\limits^\text{3 \emph{factors}}\\ &=x\cdot y\cdot x\cdot y\cdot x\cdot y\\ &=\mathop{x\cdot x\cdot x}\limits^\text{3 \emph{factors}}\cdot\mathop{y\cdot y\cdot y}\limits^\text{3 \emph{factors}}\\ &=x^3\cdot y^3 \end{align*}$

Warning Triangle with Exclamation Point Warning! This does not work when there is a + or – in the ( )! $\textcolor{red}{(xy)^2=x^2y^2}$ but $\textcolor{red}{(x+y)^2 \neq x^2+y^2}$ . More on this in the next section.

Power of a Product Rule of Exponents

For any real numbers $a$ and $b$ and any integer $n$ the power of a product rule of exponents states that

$\left( ab \right)^n=a^nb^n$

Examples for finding the power of a product

a. $\left(ab^2\right)^3$

First use the power of product rule: $\left(ab^2\right)^3=a^3\left(b^2\right)^3$

Now we apply the power rule of exponents:

$a^3\left(b^2\right)^3=a^3b^{2\cdot 3} = a^3b^6$

b. $\left(2t\right)^{15}$

$\left(2t\right)^{15}=2^{15}t^{15}=32768t^{15}$ (Using a calculator, press 2 ^ 15)

c. $\left(-2w^3 \right)^3$

$\left(-2w^3 \right)^3=(-2)^3w^{3 \cdot 3}=(-2)^3w^{9}=-8w^9$

d. $\dfrac{1}{\left(-3z\right)^4}$

$\dfrac{1}{\left(-3z\right)^4}=\dfrac{1}{\left(-3\right)^4z^4}=\dfrac{1}{81z^4}$

e. $\left(p^{-2}q^2\right)^7$

$\left(p^{-2}q^2\right)^7=p^{-2 \cdot 7}q^{2 \cdot 7}=p^{-14}q^{14}$

But note we have a negative exponent here so putting that into the denominator, we end up with $\dfrac{q^{14}}{p^{14}}$ .

Try It Now 7

a. $\left(g^2h^3\right)^5$

b. $\left(5x\right)^{3}$

c. $\left(-3y^5 \right)^3$

d. $\dfrac{1}{\left(a^6b^7\right)^2}$

e. $\left(a^{3}b^{-2}\right)^4$

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 part e in the examples above.

$\left(p^{-2}q^2\right)^7=\dfrac{q^{14}}{p^{14}}$

What if we started by rewriting the original problem using the Negative Rule of Exponents:

$\left(p^{-2}q^2\right)^7=\left( \dfrac{q^2}{p^2} \right)^7 = \dfrac{q^{14}}{p^{14}}$

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

$\begin{array}{l} {\left( {{p^{ - 2}}{q^2}} \right)^7} = {\left( {\dfrac{{{q^2}}}{{{p^2}}}} \right)^7}\\ \quad \quad \quad \;\;\; = \dfrac{{{{\left( {{q^2}} \right)}^7}}}{{{{\left( {{p^2}} \right)}^7}}}\\ \quad \quad \quad \;\;\; = \dfrac{{{q^{2 \cdot 7}}}}{{{p^{2 \cdot 7}}}}\\ \quad \quad \quad \;\;\; = \dfrac{{{q^{14}}}}{{{p^{14}}}} \end{array}$

The Power of a Quotient Rule of Exponents

For any real number $a$ and nonzero real number $b$ and any integer $n$ , the power of a quotient rule of exponents states that

$\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}$

Examples using the power of a quotient rule

a. $\left( \dfrac{4}{z^{11}} \right)^3}$

$\left( \dfrac{4}{z^{11}}\right)^3=\dfrac{4^3}{\left( z^{11} \right)^3}=\dfrac{64}{z^{33}}$

b. ${\left( {\dfrac{p}{q^3}} \right)^6}$

${\left( {\dfrac{p}{{{q^3}}}}} \right)^6}=\dfrac{p^6}{\left( q^3 \right)^6}=\dfrac{p^6}{q^{18}}$

c. ${\left( {\dfrac{-1}{{{t^{2}}}}} \right)^{27}$

$\left( \dfrac{-1}{t^2} \right)^{27}=\dfrac {\left(-1\right)^{27}}{\left(t^2\right)^{27}}=\dfrac{-1}{t^{54}}$

d. ${\left( j^3k^{-2} \right)^4}$

$\left( j^3k^{-2} \right)^4=\left( j^3 \right)^4 \left( k^{-2} \right) ^4=\dfrac{j^{12}}{k^8}$

e. $\left(m^{-2}n^{-2} \right)^3$

$\left(m^{-2}n^{-2} \right)^3}=\left(m^{-2}\right)^3 \left(n^{-2} \right) ^3=\dfrac{1}{m^6n^6}$

Try it Now 8

a. ${\left( {\dfrac{b^5}{{{c}}}} \right)^3}$

b. ${\left( {\dfrac{5}{{{u^8}}}}} \right)^4}$

c. ${\left( {\dfrac{-1}{{{w^{3}}}}} \right)^{35}$

d. ${\left( p^{-4}q^{3} \right)^8}$

e. ${\left(c^{-5}d^{-3} \right)^4}$

Simplifying Exponential Expressions

Recall that to simplify an expression means to rewrite it by combing 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.

Examples simplifying exponential expressions

Simplify each expression and write the answer with positive exponents only (assume variables represent nonzero real numbers.)

a. $\left(6m^2n^{-1}\right)^3$

Apply Power of a Product: $\left(6m^2n^{-1}\right)^3=6^3 \left(m^2\right)^3\left(n^{-1}\right)^3$
Using Power Rule of exponents: $6^3 \left(m^2\right)^3\left(n^{-1}\right)^3=216m^{2 \cdot 3}n^{-1 \cdot 3}=216m^6n^{-3}$
Now apply the Negative Exponent Rule: $216m^6n^{-3}=\dfrac{216m^6}{n^3}$

b. ${17}^5 \cdot {17}^{-4} \cdot {17}^{-3}$

Product Rule: ${17}^5 \cdot {17}^{-4} \cdot {17}^{-3}=17^{5+(-4)+(-3)}=17^{-2}$
Negative Exponent Rule: $17^{-2}=\dfrac{1}{17^2}=\dfrac{1}{289}$

c. $\left( \dfrac {u^{-1}v}{v^{-1}} \right) ^2$

Power of a product and quotient: $\left( \dfrac {u^{-1}v}{v^{-1}} \right) ^2 = \dfrac {\left( u^{-1} \right) ^2 v^2}{\left( v^{-1} \right)^{2}}$
Power Rule: $\dfrac {\left(u^{-1}\right)^2v^2}{\left(v^{-1}\right)^{2}}=\dfrac{u^{-1 \cdot 2}v^2}{v^{-1 \cdot 2}}=\dfrac{u^{-2}v^2}{v^{-2}}$
Negative Exponent Rule: $\dfrac{u^{-2}v^2}{v^{-2}}=\dfrac{v^{2}v^2}{u^{2}}$
Product Rule: $\dfrac{v^{2}v^2}{u^{2}}=\dfrac{v^4}{u^2}$

d. $\left( -2a^3b^{-1} \right) \left( 5a^{-2}b^2 \right)$

Multiply constants and using the product rule:

$\left( -2a^3b^{-1} \right) \left( 5a^{-2}b^2 \right)=-2 \cdot 5 a^{3+(-2)} \cdot b^{-1+2}=-10ab$

e. $\left(x^2 \sqrt{2} \right)^4 \left(x^2 \sqrt{2} \right) ^{-4}$

Use the product rule: $\left(x^2 \sqrt{2} \right)^4 \left(x^2 \sqrt{2} \right) ^{-4}=\left(x^2\sqrt{2}\right)^{4+(-4)}=\left(x^2 \sqrt{2} \right) ^{0}$
Use zero exponent rule: $\left(x^2 \sqrt{2} \right) ^{0}=1$

f. $\dfrac {\left(3w^2\right)^5}{\left(6w^{-2} \right)^2}$

Notice that the 6 could be written as $2 \cdot 3$ and then use power rule: $\dfrac {\left(3w^2\right)^5}{\left(6w^{-2} \right)^2}=\dfrac {3^5w^{10}}{2^2 3^2 w^{-4}}$
Now the Quotient Rule: $\dfrac {3^5 w^{10}} {2^2 3^2 w^{-4}}=\dfrac {3^{5-2} w^{10-(-4)}}{2^2}=\dfrac {3^{3}w^{14}}{2^2}=\dfrac {27w^{14}}{4}$

Try It Now 9

Simplify each expression and write the answer with positive exponents only (assume variables represent nonzero real numbers.)

a. $\left(2uv^{-2}\right)^{-3}$

b. ${x}^8 \cdot {x}^{-12}$

c. $\left( \dfrac {p^{2}q^{-3}}{q^{-1}} \right) ^2$

d. $\left( -9r^{-5}s^{3} \right) \left( 3r^{6}s^{-4} \right)$

e. $\left(\dfrac{4}{9} tw^{-2} \right)^{-3} \left(\dfrac{4}{9}tw^{-2} \right) ^{3}$

f. $\dfrac {\left(2h^2k\right)^4}{\left(7h^{-1}k^2 \right)^2}$

Using Scientific Notation

Recall at the beginning of the section that we found the number $1.3 \times 10 ^ {13}$ $1.3 \times 10^{13}$ 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 small large 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.

$2.\underbrace{\decposl{7}\decposl{8}\decposl{0}\decposl{4}\decposl{1}\decposl{8.}}_{\textsf{6 places}}$

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.

$2.780418\times10^6$

We often round to a few significant digits, which means the number of digits once a number is in scientific notation. $2.78 \times 10^6$ would be 3 significant digits, $2.8 \times 10^6$ would be 2 significant digits.

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.

$0.\underbrace{\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{4.}}_{\textsf{13 places}} 7$

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.

$4.7\times10^{-13}$

This number has 2 significant digits.

Scientific Notation

A number is written in scientific notation if it is written in the form $a \times 10^n$ , where $1 \leq a < 10$ and $n$ is an integer.

The number of digits in $a$ is called the significant digits.

Examples of Converting Standard Notation to Scientific Notation (and rounded to 4 significant digits)

a. The gross domestic product (GDP) of the US in 2019 was 21,440,000,000,000 US Dollars.

Let’s see how many decimal places we have to move to get one digit to the left of the decimal:

$2.\underbrace{\decposl{1}\decposl{4}\decposl{4}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0.}}_{\textsf{13 places}}$

It’s 13 so it’s $2.144 \times 10^{13}$ .

b. The gross national product (GNP) of India in 2019 was 145,229,310,000,000 Indian Rupees.

$1.\underbrace{\decposl{4}\decposl{5}\decposl{2}\decposl{2}\decposl{9}\decposl{3}\decposl{1}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0.}}_{\textsf{14 places}}$

It’s 14 so it’s $1.4522931 \times 10^{14}$ . But if we round to 4 significant digits, it will be $1.452 \times 10^{14}$ .

c. The Uzbekistan So’m is worth 0.00009868 US Dollars.

$0.\underbrace{\deci{0}\deci{0}\deci{0}\deci{0}\deci{9.}}_{\textsf{5 places}} 868$

We went 5 places to the right, so our exponent is ${-5}$ and the number in scientific notation becomes $9.868 \times 10^{-5}$

d. The chances of winning a PowerBall lottery is about 0.00000000342466.

$0.\underbrace{\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{3.}}_{\textsf{9 places}} 42466$

9 places to the right, so our exponent is ${-9}$ and our value is $3.42466 \times 10^{-9}$ , but if we round off to 4 significant digits, it would be $3.425 \times 10^{-9}$ Note that it rounds up to 5 because the next digit is 5 or greater.

If the given number is greater than 1, as in examples a–b the exponent of 10 is positive; and if the number is less than 1, as in examples c-d, the exponent is negative.

Try it Now 10

a. The gross national product (GNP) of the US in 2019 was 22,040,000,000,000 US Dollars.

b. The gross national product (GDP) of Japan in 2019 was 553,723,600,000,000 Japanese Yen.

c. The South Korean Won is worth 0.000823 US Dollars.

d. The chances of being killed in a shark attack is 0.0000000037864.

Converting from Scientific to Standard Notation

To convert a number in scientific notation to standard notation, simply reverse the process. Move the decimal $n$ places to the right if $n$ is positive or $n$ places to the left if $n$ is negative and add zeros as needed. Remember, if $n$ is positive, the value of the number is greater than 1, and if $n$ is negative, the value of the number is less than one.
Calculators will often given answers in scientific notation especially any number less than 0.001 or larger than 9,999,999,999. In the calculator, we often see the $a$ part of the number and then an E and then the exponent on 10. You will not see the 10. For example, you will see 4.3245318E12 to indicate $4.3245318 \times 10^{12}$ or 1.27915E-5 to indicate $1.27915 \times 10^{-5}$ .

Examples on Converting Scientific Notation to Standard Notation

a. $3.547\times10^{14}$

Move the decimal to the right 14 times filling in 0’s where you need them:

$3.\underbrace{\deci{5}\deci{4}\deci{7}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0.}}_{\textsf{14 places}}$

= 354,700,000,000,000 (I filled in the commas to see this easier.)

b. -2E6

This is calculator for $-2 \times 10^6$ and the negative here just stays in the front, and we move the decimal place to the right 6 times.

$-2.\underbrace{\deci{0}\deci{0}\deci{0}\deci{0}\deci{0}\deci{0.}}_{\textsf{6 places}}$

= -2,000,000

c. 7.91E-7

Again, we have the calculator output style here. So this means we move the decimal point 7 places to the left since it’s $10^{-7}$

$0.\underbrace{\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{7.}}_{\textsf{7 places}}91$

=0.000000791

d. $-8.05\times10^{-12}$

Since our exponent is negative, this means 12 places to the left. The negative on the 8 just goes to the front.

$-0.\underbrace{\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{0}\decposl{8.}}_{\textsf{12 places}}05$

=-0.00000000000805

Try it Now 11

a. $7.03\times10^{5}$

b. -8.16E11

c. 8E6

d. $-3.9\times10^{-13}$

Try it Now Answers

1. $k^{15}$
2. $\left( \dfrac{2}{y} \right)^5$
3. $t^8$
1. $c^7$
2. $-27$
3. $uv$
1. $(3y)^{24}$
2. $t^{35}$
3. $g^{16}$ (the negative disappears because it’s inside the parenthesis with an even exponent)
1. 1
2. $\dfrac{1}{2}$
3. 1
1. $\dfrac{1}{x^6}$
2. $\dfrac{1}{t^4}$
3. $\dfrac{2}{5k^3}$
1. $\dfrac{1}{t^3}$
2. $\dfrac{1}{x}$
1. $g^{10}h^{15}$
2. $125x^3$
3. $-27y^{15}$
4. $\dfrac{1}{a^{12}b^{14}}$
5. $\dfrac{a^{12}}{b^8}$
1. $\dfrac{b^{15}}{c^3}$
2. $\dfrac{625}{u^{32}}$
3. $-\dfrac{1}{w^{105}}$
4. $\dfrac{q^{24}}{p^{32}}$
5. $\dfrac{1}{c^{20}d^{12}}$
1. $\dfrac{v^6}{8u^3}$
2. $\dfrac{1}{x^4}$
3. $\dfrac{p^4}{q^4}$
4. $\dfrac{-27r}{s}$
5. 1
6. $\dfrac{16h^{10}}{49}$