Laws of Exponents and Scientific Notation Math 8 Review

1.2 Exponents and Scientific Notation

Topics covered in this section are:

• Employ the product rule of exponents.
• Apply the quotient rule of exponents.
• Use the power rule of exponents.
• Use the zilch exponent rule of exponents.
• Employ the negative rule of exponents.
• Find the power of a product and a quotient.
• Simplify exponential expressions.
• Use scientific notation.

Mathematicians, scientists, and economists normally run into very large and very small numbers. But it may not be obvious how common such figures are in everyday life. For example, a pixel is the smallest unit of light that tin can be perceived and recorded by a digital camera. A item camera might tape an image that is $2,048$ pixels past $i,536$ pixels, which is a very high resolution moving picture. It tin too perceive a colour depth (gradations in colors) of upwards to $48$ $.25 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 ($iii,600$-second) digital motion picture is and then an extremely large number.

Using a computer, nosotros enter $2,048 \times ane,536 \times 48 \times 24 \times 3,600$ and printing ENTER. The computer displays $1.304596316E13$. What does this mean? The "E13" portion of the upshot represents the exponent $13$ of ten, and so in that location are a maximum of approximately $1.iii \times 10^{13}$ bits of data in that one-hour pic. In this section, we review rules of exponents first and so apply them to calculations involving very large or minor numbers.

1.two.1 Using the Production Rule of Exponents

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

$x^{3} \cdot 10^{iv}$	$=$	$3$ factors $ \ \ \ \ \ \ \ \ $ $4$ factors $10 \cdot x \cdot 10 \cdot x \cdot x \cdot x \cdot 10$
	$=$	$ \ \ \ \ \ 7 $ factors $ \ \ \ \ \ \ \ \ \ \ $ $ten \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x$
	$=$	$x^{7}$

The result is that $x^{3} \cdot x^{four} = ten^{3+4} = 10^{7}$.

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 of operations, we write the result with the common base and add the exponents. This is theproduct rule of exponents.

$a^{m} \cdot a^{northward} = a^{m+north}$

At present consider an example with existent numbers.

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

Find that the exponent of the product is the sum of the exponents of the terms. In other words,

We can e'er bank check that this is true by simplifying each exponential expression. We find that $2^{iii}$ is $viii$, $two^{4}$ is $16$, and $2^{7}$ is $128$. The product $eight \cdot 16$ equals $128$, so the relationship is true. We can utilise the product dominion of exponents to simplify expressions that are a production of ii numbers or expressions with the same base only different exponents.

THE PRODUCT Rule OF EXPONENTS

For any real number $a$ and natural numbers $grand$ and $due north$, the product rule of exponents states that
$a^{m} \cdot a^{n} = a^{m+n}$

Instance 1

Write each of the following products with a single base of operations. Do non simplify farther.

• $t^{five} \cdot t^{3}$
• $(-3)^{5} \cdot (-iii)$
• $ten^{2} \cdot x^{5} \cdot x^{iii}$

Solution

• $t^{5} \cdot t^{three} = t^{5+3} = t^{8}$
• $(-3)^{5} \cdot (-3) = (-iii)^{5} \cdot (-three)^{1} = (-3)^{5+1} = (-3)^{vi}$
• $x^{two} \cdot x^{v} \cdot 10^{three}$
  At get-go, it may announced that we cannot simplify a product of three factors. However, using the associative belongings of multiplication, begin past simplifying the first ii.
  $x^{2} \cdot x^{5} \cdot x^{3} = (x^{2} \cdot x^{5}) \cdot x^{3} = (ten^{2+5}) \cdot x^{3} = 10^{seven} \cdot 10^{iii} = x^{7+3} = ten^{x}$
  Notice we get the same result by adding the 3 exponents in i step.
  $x^{2} \cdot x^{five} \cdot 10^{3} = x^{2+v+3} = x^{ten}$

1.2.2 Using the Quotient Rule of Exponents

Thequotient rule of exponents allows u.s. to simplify an expression that divides two numbers with the aforementioned base of operations merely dissimilar exponents. In a similar way to the production rule, nosotros can simplify an expression such every bit $\frac {y^{thousand}} {y^{due north}}$, where $m>n$. Consider the case $\frac {y^{9}} {y^{5}}$.Perform the division past canceling common factors.

$\begin{align*} \frac {y^{9}} {y^{5}} &= \frac {y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y} {y \cdot y \cdot y \cdot y \cdot y} \\ &= \frac { \abolish{y} \cdot \cancel{y} \cdot \abolish{y} \cdot \abolish{y} \cdot \cancel{y} \cdot y \cdot y \cdot y \cdot y} {\cancel{y} \cdot \abolish{y} \cdot \cancel{y} \cdot \cancel{y} \cdot \abolish{y}} \\ &= \frac {y \cdot y \cdot y \cdot y}{i} \\ &=y^{iv} \end{align*}$

Observe that the exponent of the caliber is the difference between the exponents of the divisor and dividend.

$\frac {a^{m}}{a^{n}} = a^{thousand-n}$

In other words, when dividing exponential expressions with the aforementioned base, nosotros write the consequence with the common base and subtract the exponents.

$\frac {y^{9}}{y^{five}} = y^{9-five} = y^{4}$

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

THE QUOTIENT Dominion OF EXPONENTS

For any existent number $a$ and natural numbers $k$ and $n$, such that $thou>n$, the caliber rule of exponents states that
$\frac {a^{chiliad}}{a^{n}} = a^{k-northward}$

Example 2

Write each of the following products with a single base of operations. Practice non simplify further.

• $\frac{(-2)^{xiv}} {(-2)^{ix}}$
• $\frac{t^{23}} {t^{fifteen}}$
• $\frac{(z \sqrt{2})^{v}} {z \sqrt{two}}$

Solution

Use the quotient rule to simplify each expression.

• $\frac{(-ii)^{fourteen}} {(-ii)^{9}} = (-ii)^{14-9} = (-2)^{5}$
• $\frac{t^{23}} {t^{15}} = t^{23-fifteen} = t^{8}$
• $\frac{(z \sqrt{2})^{5}} {z \sqrt{2}} = (z \sqrt{2})^{5-1} = (z \sqrt{2})^{4}$

1.2.three Using the Power Rule of Exponents

Suppose an exponential expression is raised to some power. Tin can we simplify the result? Yes. To do this, nosotros use theability dominion of exponents. Consider the expression $(x^{2})^{3}$. The expression within the parentheses is multiplied twice because it has an exponent of $2$. Then the result is multiplied three times considering the unabridged expression has an exponent of $three$.

$(ten^{2})^{3}$	$=$	$3$ factors $(ten^{ii}) \cdot (ten^{2}) \cdot (x^{2})$
	$=$	$3$ factors $(ten \cdot ten) \ (x \cdot x) \ (x \cdot 10)$ $2$ factors $2$ factors $two$ factors
	$=$	$ 10 \cdot x \cdot x \cdot 10 \cdot 10 \cdot ten$
	$=$	$x^{half dozen}$

The exponent of the respond is the product of the exponents: $(x^{2})^{3} = x^{2 \cdot iii} = x^{six}$. 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.

$(a^{grand})^{north} = a^{m \cdot n}$

Be conscientious to distinguish between uses of the product rule and the power rule. When using the product rule, dissimilar 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.

		Ability	Rule					Product	Rule
$5^{iii} \cdot five^{4}$	$=$	$5^{3+4}$	$=$	$five^{7}$	but	$(five^{3})^{four}$	$=$	$v^{three \cdot four}$	$=$	$five^{12}$
$x^{v} \cdot ten^{2}$	$=$	$x^{5+2}$	$=$	$x^{vii}$	just	$(x^{5})^{2}$	$=$	$ten^{5 \cdot 2}$	$=$	$x^{10}$
$(3a)^{7} \cdot (3a)^{ten}$	$=$	$(3a)^{7+10}$	$=$	$(3a)^{17}$	but	$((3a)^{vii})^{10}$	$=$	$(3a)^{7 \cdot x}$	$=$	$(3a)^{70}$

THE POWER RULE OF EXPONENTS

For any real number $a$ and positive integers $m$ and $n$, the power rule states that
$(a^{m})^{n} = a^{g \cdot north}$

Example 3

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

• $(x^{ii})^{vii}$
• $((2t)^{v})^{3}$
• $((-3)^{5})^{eleven}$

Solution

Use the power rule to simplify each expression.

• $(x^{2})^{seven} = x^{ii \cdot vii} = ten^{14}$
• $((2t)^{5})^{3} = (2t)^{5 \cdot 3} = (2t)^{xv}$
• $((-3)^{5})^{eleven} = (-3)^{five \cdot xi} = (-3)^{55}$

1.2.4 Using the Zero Exponent Dominion of Exponents

Render to the quotient dominion. We made the status that $m>northward$ so that the difference $m-n$ would never be nothing or negative. What would happen if $1000=n$? In this case, we would use thecipher exponent rule of exponents to simplify the expression to $ane$. To see how this is washed, permit us begin with an example.

$\frac {t^{8}} {t^{8}} = \frac { \cancel{t^{8}}} { \cancel{t^{8}}} = i$

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

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

If we equate the two answers, the issue is $t^{0}=1$. This is truthful 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 afterward in more than advanced courses, only for now, we will consider the value to be undefined.

THE Null EXPONENT RULE OF EXPONENTS

For whatsoever nonzero real number $a$, the zero exponent rule of exponents states that
$a^{0}=one$

Case 4

Simplify each expression using the nil exponent dominion of exponents.

• $\frac{c^{three}} {c^{3}}$
• $\frac{-3x^{5}} {x^{5}}$
• $\frac{(j^{ii}k)^{4}} {(j^{ii}thousand) \cdot (j^{ii}k)^{3}}$
• $\frac {5(rs^{2})^{two}} {(rs^{2})^{2}}$

Solution

Utilize the cypher exponent and other rules to simplify each expression.

• $\brainstorm{align*} \frac{c^{iii}} {c^{3}} &= c^{three-three} \\ &= c^{0} \\ &= 1 \\ \ \ \ \ \\ \ \ \ \ \\ \end{align*}$
• $\begin{align*} \frac{-3x^{5}} {x^{five}} &= -3 \cdot \frac{ x^{five}} {10^{5}} \\ &= -3 \cdot x^{5-5} \\ &= -3 \cdot x^{0} \\ &= -iii \cdot 1 \\ &= -3 \\ \ \ \ \ \\ \ \ \ \ \\ \end{align*}$
• $\begin{marshal*} \frac{(j^{two}chiliad)^{4}} {(j^{two}one thousand) \cdot (j^{2}k)^{iii}} &= \frac{(j^{2}k)^{four}} {(j^{two}k)^{one+3}} \\ &= \frac{(j^{two}m)^{4}} {(j^{two}thousand)^{4}} \\ &= (j^{2}k)^{4-4} \\ &= (j^{2}1000)^{0} \\ &= i \\ \ \ \ \ \\ \ \ \ \ \\ \end{marshal*}$
• $\brainstorm{align*} \frac {five(rs^{two})^{ii}} {(rs^{2})^{2}} &= v(rs^{2})^{2-2} \\ &= 5(rs^{2})^{0} \\ &= 5 \cdot 1 \\ &= v \\ \ \ \ \ \\ \ \ \ \ \\ \finish{align*}$

1.2.5 Using the Negative Rule of Exponents

Some other useful result occurs if we relax the condition that $thou>northward$ in the quotient rule fifty-fifty further. For example, tin can we simplify $\frac { h^{iii}} {h^{5}}$? When $m<n$—that is, where the departure $m-n$ is negative—we can use thenegative dominion of exponents to simplify the expression to its reciprocal.

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

$ \begin {align*} \frac { h^{iii}} {h^{v}} &= \frac {h \cdot h \cdot h} {h \cdot h \cdot h \cdot h \cdot h} \\ &= \frac { \cancel{h} \cdot \cancel{h} \cdot \cancel{h}} {\cancel{h} \cdot \abolish{h} \cdot \abolish{h} \cdot h \cdot h} \\ &= \frac {i} {h \cdot h} \\ & = \frac {1} { h^{2}} \end{marshal*}$

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

$ \begin{align*} \frac {h^{3}} {h^{5}} &= h^{iii-5} \\ &= h^{-2} \end{marshal*} $

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

A factor with a negative exponent becomes the same factor with a positive exponent if information technology is moved beyond the fraction bar—from numerator to denominator or vice versa.

$a^{-n} = \frac {1} {a^{n}}$ and $a^{north} = \frac {one} {a^{-n}}$

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 production and caliber rules and all of the rules we will look at soon hold for whatever integer $northward$.

THE NEGATIVE RULE OF EXPONENTS

For any nonzero real number $a$ and natural number $due north$, the negative rule of exponents states that
$a^{-n} = \frac {one} {a^{n}}$

Example five

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

• $\frac {θ^{3}} {θ^{10}}$
• $\frac {z^{2} \cdot z} {z^{four}} $
• $\frac {(-5t^{3})^{4}} {(-5t^{3})^{eight}}$

Solution

• $\frac {θ^{3}} {θ^{10}} = θ^{3-10} = θ^{-7} = \frac{one} {θ^{vii}}$
• $\frac{z^{ii} \cdot z} {z^{4}} = \frac {z^{2+1}} {z^{4}} = \frac {z^{3}} {z^{four}} =z^{three-4} = z^{-1} = \frac {1}{z}$
• $\frac {(-5t^{3})^{iv}} {(-5t^{three})^{8}} = (-5t^{3})^{four-viii} = (-5t^{3})^{-4} = \frac {1} {(-5t^{three})^{4}}$

Example 6

Write each of the following products with a single base. Do non simplify further. Write answers with positive exponents.

• $b^{two} \cdot b^{-viii}$
• $(-x)^{5} \cdot (-x)^{-v}$
• $\frac{-7z} {(-7z)^{5}}$

Solution

• $b^{2} \cdot b^{-8} = b^{2-8} = b^{-6} = \frac{1} {b^{half-dozen}}$
• $(-x)^{v} \cdot (-ten)^{-five} = (-10)^{5-5} = (-x)^{0} = one$
• $\frac{-7z} {(-7z)^{v}} = (-7z)^{one-5} = (-7z)^{-iv} = \frac {1} {(-7z)^{iv}}$

ane.2.6a Finding the Ability of a Product

To simplify the power of a product of 2 exponential expressions, we tin use theability 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 $(pq)^{3}$. Nosotros begin by using the associative and commutative properties of multiplication to regroup the factors.

	3 factors
$(pq)^{3}$	$= (pq) \cdot (pq) \cdot (pq)$
	$= p \cdot q \cdot p \cdot q \cdot p \cdot q$
	3 factors $\ \ \ \ \ \ \ $ three factors
	$= p \cdot p \cdot p \cdot q \cdot q \cdot q$
	$=p^{3} \cdot q^{3}$

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

THE 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
$(ab)^{n}=a^{due north}b^{north}$

Example 7

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

• $(ab^{2})^{3}$
• $(2t)^{15}$
• $(-2w^{three})^{iii}$
• $\frac{1} {(-7z)^{4}}$
• $(e^{-2}f^{2})^{seven}$

Solution

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

• $(ab^{2})^{3}=(a)^{iii} \cdot (b^{2})^{3} = a^{1 \cdot iii} \cdot b^{ii \cdot three} = a^{3}b^{six}$
• $(2t)^{15} = (ii)^{xv} \cdot (t)^{15} = 2^{xv}t^{15} = 32,768t^{15}$
• $(-2w^{3})^{iii} = (-two)^{3} \cdot (w^{3})^{iii} = -8 \cdot west^{three \cdot 3} = -8w^{nine}$
• $\frac{1} {(-7z)^{4}} = \frac{1} {(seven)^{4} \cdot (z)^{4}} = \frac {1} {2,401z^{4}}$
• $(due east^{-2}f^{two})^{7} = (e^{-2})^{vii} \cdot (f^{2})^{7} = e^{-ii \cdot vii} \cdot f^{2 \cdot 7} = due east^{-14}f^{fourteen} = \frac {f^{14}} {due east^{14}}$

1.ii.6b Finding the Power of a Quotient

To simplify the power of a quotient of ii expressions, we can use thepower of a quotient rule, which states that the power of a quotient of factors is the caliber of the powers of the factors. For example, let's look at the following example.

$(e^{-2}f^{2})^{seven} = \frac {f^{14}} {e^{xiv}}$

Let's rewrite the original problem differently and wait at the result.

$(e^{-ii}f^{2})^{seven} = (\frac {f^{ii}} {east^{2}})^{7}$
$ \ \ \ \ \ \ \ \ \ = \frac {f^{14}} {e^{14}}$

Information technology appears from the last ii steps that nosotros can use the power of a production rule as a power of a quotient rule.

$ \begin{align*} (e^{-2}f^{2})^{vii} &= (\frac {f^{ii}} {e^{ii}})^{7} \\ &= \frac{(f^{two})^{7}} {(due east^{2})^{7}} \\ &= \frac {f^{2 \cdot 7}} {e^{2 \cdot vii}} \\ &= {f^{14}} {e^{fourteen}} \finish{align*}$

THE POWER OF A QUOTIENT Rule OF EXPONENTS

For any real numbers $a$ and $b$ and whatsoever integer $due north$, the power of a quotient rule of exponents states that
$(\frac{a}{b})^{n} = \frac{a^{northward}} {b^{n}} $

Instance viii

Simplify each of the post-obit quotients as much equally possible using the ability of a quotient dominion. Write answers with positive exponents.

• $(\frac {four}{z^{11}})^{3}$
• $(\frac {p}{q^{3}})^{6}$
• $(\frac {-1}{t^{2}})^{27}$
• $(j^{three}thou^{-2})^{four}$
• $(thou^{-2}north^{-2})^{3}$

Solution

• $(\frac {4}{z^{eleven}})^{iii} = \frac{(4)^{3}} {(z^{eleven})^{iii}} = \frac{64}{z^{11 \cdot 3}} = \frac{64}{z^{33}}$
• $(\frac {p}{q^{3}})^{half-dozen} = \frac{(p)^{6}} {(q^{3})^{6}} = \frac{p^{1 \cdot 6}}{q^{3 \cdot half-dozen}} = \frac{p^{half-dozen}} {q^{eighteen}}$
• $(\frac {-1}{t^{2}})^{27} = = \frac{(-i)^{27}} {(t^{ii})^{27}} = \frac{-1}{t^{2 \cdot 27}} = – \frac{1} {t^{54}}$
• $(j^{3}k^{-ii})^{4} = (\frac{j^{3}} {one thousand^{two}})^{iv} = \frac{(j^{three})^{4}} {(1000^{2})^{four}} = \frac {j^{3 \cdot 4}} {k^{2 \cdot 4}} = \frac{j^{12}} {one thousand^{8}}$
• $(grand^{-2}n^{-2})^{3} = (\frac{1}{g^{2}n^{2}})^{3} = \frac{(1)^{3}} {(m^{ii}n^{2})^{3}} = \frac{i}{(m^{2})^{3}(n^{2})^{three}} = \frac{1}{chiliad^{2 \cdot 3} \cdot n^{2 \cdot 3}} = \frac{1}{k^{six}n^{vi}}$

one.2.7 Simplifying Exponential Expressions

Recall that to simplify an expression means to rewrite information technology 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.

Example 9

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

• $(6m^{2}n^{-ane})^{three}$
• $17^{5} \cdot 17^{-4} \cdot 17^{-three}$
• $(\frac{u^{-1}v}{v^{-1}})^{2}$
• $(-2a^{-iii}b^{-ane})(5a^{-2}b^{2})$
• $(x^{2} \sqrt{2})^{four}(10^{ii} \sqrt{two})^{-4}$
• $\frac{(3w^{2})^{5}} {(6w^{-2})^{2}}$

Solution

Function 1

$(6m^{2}due north^{-ane})^{three}$	$=(6)^{3}(m^{2})^{3}(n^{-one})^{3}$	The power of a product dominion
	$=6^{three}m^{two \cdot three}due north^{-1 \cdot 3}$	The power rule
	$=216m^{6}n^{-three}$	Simplify.
	$=\frac {216m^{6}} {n^iii}$	The negative exponent dominion

Part ii

$17^{five} \cdot 17^{-four} \cdot 17^{-3}$	$=17^{5-4-3}$	The production rule
	$=17^{-two}$	Simplify.
	$=\frac{one}{17^{two}}$ or $\frac {1}{289} $	The negative exponent dominion

Part iii

$(\frac{u^{-1}five}{v^{-1}})^{2}$	$=\frac{(u^{-1}five)^{2}}{(v^{-one})^{ii}}$	The power of a quotient rule
	$=\frac{(u^{-2}v^{2}}{v^{-2}}$	The ability of a product rule
	$=u^{-2}v^{two-(-2)}$	The quotient dominion
	$=u^{-2}five^{4}$	Simplify.
	$=\frac{v^{4}}{u^{2}}$	The negative exponent rule

Part 4

$(-2a^{3}b^{-1})(5a^{-ii}b^{2})$	$=-2 \cdot 5 \cdot a^{3} \cdot a^{-2} \cdot b^{-one} \cdot b^{ii}$	Commutative and associative laws of multiplication
	$=-ten \cdot a^{3-2} \cdot b^{-1+2}$	The production rule
	$=-10ab$	Simplify.

Part 5

$(x^{2} \sqrt{2})^{4}(x^{two} \sqrt{ii})^{-4}$	$=(x^{ii} \sqrt{ii})^{4-iv}$	The product rule
	$=(10^{2} \sqrt{2})^{0}$	Simplify.
	$=i$	The nix exponent dominion

Office 6

$\frac{(3w^{two})^{5}}{(6w^{-2})^{2}}$	$=\frac{(3)^{5} \cdot (w^{ii})^{5}} {(half-dozen)^{2} \cdot (w^{-two})^{2}}$	The power of a product rule
	$=\frac{3^{5}west^{ii \cdot 5}} {6^{2}w^{-ii \cdot ii}}$	The power dominion
	$=\frac{243w^{ten}} {36w^{-four}}$	Simplify.
	$=\frac{27w^{10-(-4)}} {iv}$	The quotient rule and reduce fraction
	$=\frac{27w^{14}} {4}$	Simplify.

1.2.8 Using Scientific Notation

Recollect at the beginning of the department that we found the number $ane.3 \times 10^{13}$ when describing bits of information in digital images. Other extreme numbers include the width of a human hair, which is about $0.00005$ 1000, and the radius of an electron, which is about $0.00000000000047$ m. How tin can we effectively piece of work read, compare, and calculate with numbers such as these?

A shorthand method of writing very minor and very large numbers is chosenscientific notation, in which we limited numbers in terms of exponents of $10$. To write a number in scientific note, move the decimal point to the correct of the first digit in the number. Write the digits as a decimal number between $1$ and $x$. Count the number of places $northward$ that you moved the decimal point. Multiply the decimal number by $10$ raised to a ability of $n$. If you moved the decimal left equally in a very big number, $n$ is positive. If you moved the decimal right as in a pocket-size large number, $n$ is negative.

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

We obtain $ii.780418$ by moving the decimal point 6 places to the left. Therefore, the exponent of $ten$ is $6$, and it is positive because we moved the decimal point to the left. This is what we should look for a large number.

$two.780418 \times 10^{vi}$

Working with pocket-size numbers is like. Take, for instance, the radius of an electron, $0.00000000000047$ thou. Perform the same series of steps as above, except motility the decimal point to the right.

Exist conscientious non to include the leading $0$ in your count. We motion the decimal bespeak $13$ places to the right, so the exponent of $x$ is $xiii$. The exponent is negative considering nosotros moved the decimal betoken to the right. This is what we should expect for a modest number.

$4.7 \times ten^{-thirteen}$

SCIENTIFIC Note

A number is written inscientific note if it is written in the form $a \times x^{north}$, where $1≤|a|<10$ and and $n$ is an integer.

Instance 10

Write each number in scientific notation.

• Distance to Andromeda Galaxy from Earth: $24,000,000,000,000,000,000,000$ thousand
• Diameter of Andromeda Galaxy: $1,300,000,000,000,000,000,000$ thousand
• Number of stars in Andromeda Galaxy: $one,000,000,000,000$
• Diameter of electron: $0.00000000000094$ m
• Probability of beingness struck past lightning in any unmarried year: $0.00000143$

Solution

$24,000,000,000,000,000,000,000$ m	$24,000,000,000,000,000,000,000$ m
	<- $22$ places
	$2.iv \times 10^{22}$ m
$one,300,000,000,000,000,000,000$ m	$ane,300,000,000,000,000,000,000$ thou
	<- $21$ places
	$1.3 \times 10^{21}$ g
$1,000,000,000,000$	$one,000,000,000,000$
	<- $12$ places
	$ane \times x^{12}$
$0.00000000000094$ 1000	$0.00000000000094$
	-> 13 places
	$9.4 \times 10^{-13}$ g
$0.00000143$	$0.00000143$
	-> half-dozen places
	$one.43 \times 10^{-6}$

Observe that, if the given number is greater than $1$, as in the first iii examples, the exponent of $x$ is positive; and if the number is less than $one$, as in the terminal three examples, the exponent is negative.

Converting from Scientific to Standard Notation

To convert a number inscientific notation to standard annotation, but contrary the process. Movement the decimal $n$ places to the right if $northward$ is positive or $n$ places to the left if $n$ is negative and add zeros as needed. Remember, if $due north$ is positive, the value of the number is greater than $1$, and if $northward$ is negative, the value of the number is less than ane.

Example xi

Catechumen each number in scientific notation to standard note.

• $3.547 \times 10^{fourteen}$
• $−2 \times 10^{vi}$
• $7.91 \times 10^{−7}$
• $−eight.05 \times 10^{−12}$

Solution

$3.547 \times 10^{14}$	$iii.54700000000000$
	-> $xiv$ places
	$354,700,000,000,000$
$-2 \times 10^{6}$	$-2.000000$
	-> $six$ places
	$-2,000,000$
$7.91 \times 10^{-7}$	$0000007.91$
	<- $7$ places
	$0.000000791$
$-8.05 \times 10^{-12}$	$-000000000008.05$
	<- $12$ places
	$-0.00000000000805$

Using Scientific Notation in Applications

Scientific notation, used with the rules of exponents, makes computing with large or small numbers much easier than doing so using standard notation. For case, suppose we are asked to calculate the number of atoms in $i$ L of water. Each water molecule contains $3$ atoms ($2$ hydrogen and $one$ oxygen). The average driblet of h2o contains around $ane.32 \times 10^{21}$ molecules of water and $1$ L of h2o holds about $1.22 \times 10^{4}$ average drops. Therefore, at that place are approximately $3 \cdot (ane.32 \times 10^{21}) \cdot (1.22 \times x^{4}) \approx 4.83 \times 10^{25}$ atoms in $ane$ 50 of water. We simply multiply the decimal terms and add the exponents. Imagine having to perform the calculation without using scientific annotation!

When performing calculations with scientific annotation, be sure to write the answer in proper scientific annotation. For case, consider the production $(7 \times ten^{4}) \cdot (v \times 10^{6}) = 35 \times 10^{10}$. The answer is not in proper scientific notation because $35$ is greater than $10$. Consider $35$ as $3.v \times 10$. That adds a ten to the exponent of the answer.

$(35) \times 10^{x} = (3.5 \times 10) \times ten^{x} = iii.5 \times (x \times 10^{10}) = three.v \times 10^{11}$

Case 12

Perform the operations and write the answer in scientific notation.

• $(8.14 \times 10^{-7}) (half-dozen.5 \times 10^{10})$
• $(4 \times ten^{5}) \div (-i.52 \times ten^{nine})$
• $(2.7 \times 10^{5}) (6.04 \times x^{13})$
• $(one.two \times 10^{8}) \div (ix.half dozen \times 10^{5})$
• $(3.33 \times 10^{4})(-1.05 \times 10^{vii})(5.62 \times 10^{5})$

Solution

$(8.fourteen \times x^{-7})(half dozen.5 \times x^{x})$	$=(8.14 \times half-dozen.5)(10^{-7} \times 10^{10})$	Commutative and associative properties of multiplication
	$=(52.91)(x^{3})$	Product rule of exponents
	$=five.291 \times 10^{4}$	Scientific notation
$(4 \times x^{five}) \div (-ane.52 \times 10^{9})$	$=(\frac {4}{-ane.52})(\frac {10^{v}} {10^{9}})$	Commutative and associative backdrop of multiplication
	$ \approx (-2.63)(10^{-4})$	Caliber rule of exponents
	$=-two.63 \times x^{-4}$	Scientific Note
$(ii.seven \times 10^{v})(half dozen.04 \times10^{13})$	$=(2.7 \times 6.04)(10^{five} \times10^{xiii})$	Commutative and associative backdrop of multiplication
	$=(16.308)(10^{18})$	Product rule of exponents
	$=ane.6308 \times ten^{xix}$	Scientific Notation
$(ane.ii \times 10^{eight}) \div (nine.half dozen \times 10^{five})$	$=(\frac{1.ii}{9.6}) (\frac{10^{8}} {10^{5}})$	Commutative and associative backdrop of multiplication
	$=(0.125)(ten^{3})$	Caliber rule of exponents
	$=1.25 \times 10^{2}$	Scientific Notation
$(3.33 \times 10^{4}) (-1.05 \times ten^{vii}) (5.62 \times ten^{5})$	$=[3.33 \times (-1.05) \times five.62] (10^{4} \times 10^{7} \times x^{5})$
	$\approx (-nineteen.65)(10^{16})$
	$=-i.965 \times x^{17}$

Example 13

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 note, 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.

Solution

The population was $308,000,000 = 3.08 \times 10 ^{8}$.
The national debt was $\$17,547,000,000,000 \approx \$1.75 \times 10^{xiii}$.
To find the amount of debt per citizen, split the national debt by the number of citizens.

$\begin{marshal*} (1.75 \times 10^{xiii}) \div (three.08 \times 10^{8}) &= (\frac{ane.75}{3.08}) \cdot (\frac{x^{13}} {10^{viii}}) \\ & \approx 0.57 \times x^{5} \\ &=five.7 \times 10^{4} \stop{align*}$

The debt per citizen at the time was most $\$5.7 \times x^{4}$, or $\$57,000$.

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Source: https://minutemath.com/college-algebra/exponents-and-scientific-notation/

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