Exponents and Scientific Notation

Learning Objectives

In this section, you will:

• 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.

Mathematicians, scientists, and economists commonly encounter very large and very small numbers. But it may not be obvious how common such figures 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 pixel, 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 2,048×1,536×48×24×3,6002,048×1,536×48×24×3,600 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 1.3×10131.3×1013 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 x3⋅x4.𝑥3⋅𝑥4. Both terms have the same base, x, but they are raised to different exponents. Expand each expression, and then rewrite the resulting expression.

x3⋅x4===x⋅x⋅x3factors⋅x⋅x⋅x⋅x4factorsx⋅x⋅x⋅x⋅x⋅x⋅x7factorsx7𝑥3⋅𝑥4=𝑥⋅𝑥⋅𝑥3factors⋅𝑥⋅𝑥⋅𝑥⋅𝑥4factors=𝑥⋅𝑥⋅𝑥⋅𝑥⋅𝑥⋅𝑥⋅𝑥7factors=𝑥7

The result is that x3⋅x4=x3+4=x7.𝑥3⋅𝑥4=𝑥3+4=𝑥7.

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.

am⋅an=am+n𝑎𝑚⋅𝑎𝑛=𝑎𝑚+𝑛

Now consider an example with real numbers.

23⋅24=23+4=2723⋅24=23+4=27

We can always check that this is true by simplifying each exponential expression. We find that 2323 is 8, 2424 is 16, and 2727 is 128. The product 8⋅168⋅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.

THE PRODUCT RULE OF EXPONENTS

For any real number a𝑎 and natural numbers m𝑚 and n,𝑛, the product rule of exponents states that

am⋅an=am+n𝑎𝑚⋅𝑎𝑛=𝑎𝑚+𝑛

EXAMPLE 1

Using the Product Rule

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

1. ⓐ t5⋅t3𝑡5⋅𝑡3
2. ⓑ (−3)5⋅(−3)(−3)5⋅(−3)
3. ⓒ x2⋅x5⋅x3𝑥2⋅𝑥5⋅𝑥3

Solution

Use the product rule to simplify each expression.

1. ⓐ t5⋅t3=t5+3=t8𝑡5⋅𝑡3=𝑡5+3=𝑡8
2. ⓑ (−3)5⋅(−3)=(−3)5⋅(−3)1=(−3)5+1=(−3)6(−3)5⋅(−3)=(−3)5⋅(−3)1=(−3)5+1=(−3)6
3. ⓒ x2⋅x5⋅x3𝑥2⋅𝑥5⋅𝑥3

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.

x2⋅x5⋅x3=(x2⋅x5)⋅x3=(x2+5)⋅x3=x7⋅x3=x7+3=x10𝑥2⋅𝑥5⋅𝑥3=(𝑥2⋅𝑥5)⋅𝑥3=(𝑥2+5)⋅𝑥3=𝑥7⋅𝑥3=𝑥7+3=𝑥10

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

x2⋅x5⋅x3=x2+5+3=x10𝑥2⋅𝑥5⋅𝑥3=𝑥2+5+3=𝑥10

TRY IT #1

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

1. ⓐ k6⋅k9𝑘6⋅𝑘9
2. ⓑ (2y)4⋅(2y)(2𝑦)4⋅(2𝑦)
3. ⓒ t3⋅t6⋅t5𝑡3⋅𝑡6⋅𝑡5

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 ymyn,𝑦𝑚𝑦𝑛, where m>n.𝑚>𝑛. Consider the example y9y5.𝑦9𝑦5. Perform the division by canceling common factors.

y9y5====y⋅y⋅y⋅y⋅y⋅y⋅y⋅y⋅yy⋅y⋅y⋅y⋅yy⋅y⋅y⋅y⋅y⋅y⋅y⋅y⋅yy⋅y⋅y⋅y⋅yy⋅y⋅y⋅y1y4𝑦9𝑦5=𝑦⋅𝑦⋅𝑦⋅𝑦⋅𝑦⋅𝑦⋅𝑦⋅𝑦⋅𝑦𝑦⋅𝑦⋅𝑦⋅𝑦⋅𝑦=𝑦⋅𝑦⋅𝑦⋅𝑦⋅𝑦⋅𝑦⋅𝑦⋅𝑦⋅𝑦𝑦⋅𝑦⋅𝑦⋅𝑦⋅𝑦=𝑦⋅𝑦⋅𝑦⋅𝑦1=𝑦4

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

aman=am−n𝑎𝑚𝑎𝑛=𝑎𝑚−𝑛

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

y9y5=y9−5=y4𝑦9𝑦5=𝑦9−5=𝑦4

For the time being, we must be aware of the condition m>n.𝑚>𝑛. Otherwise, the difference m−n𝑚−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.

THE QUOTIENT RULE OF EXPONENTS

For any real number a𝑎 and natural numbers m𝑚 and n,𝑛, such that m>n,𝑚>𝑛, the quotient rule of exponents states that

aman=am−n𝑎𝑚𝑎𝑛=𝑎𝑚−𝑛

EXAMPLE 2

Using the Quotient Rule

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

1. ⓐ (−2)14(−2)9(−2)14(−2)9
2. ⓑ t23t15𝑡23𝑡15
3. ⓒ (z2√)5z2√(𝑧2)5𝑧2

Solution

Use the quotient rule to simplify each expression.

1. ⓐ (−2)14(−2)9=(−2)14−9=(−2)5(−2)14(−2)9=(−2)14−9=(−2)5
2. ⓑ t23t15=t23−15=t8𝑡23𝑡15=𝑡23−15=𝑡8
3. ⓒ (z2√)5z2√=(z2–√)5−1=(z2–√)4(𝑧2)5𝑧2=(𝑧2)5−1=(𝑧2)4

TRY IT #2

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

1. ⓐ s75s68𝑠75𝑠68
2. ⓑ (−3)6−3(−3)6−3
3. ⓒ (ef2)5(ef2)3(𝑒𝑓2)5(𝑒𝑓2)3

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 (x2)3.(𝑥2)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.

(x2)3====(x2)⋅(x2)⋅(x2)3factors⎛⎝x⋅x2factors⎞⎠⋅⎛⎝x⋅x2factors⎞⎠⋅⎛⎝x⋅x2factors⎞⎠3factorsx⋅x⋅x⋅x⋅x⋅xx6(𝑥2)3=(𝑥2)⋅(𝑥2)⋅(𝑥2)3factors=(𝑥⋅𝑥︷2factors)⋅(𝑥⋅𝑥︷2factors)⋅(𝑥⋅𝑥︷2factors)3factors=𝑥⋅𝑥⋅𝑥⋅𝑥⋅𝑥⋅𝑥=𝑥6

The exponent of the answer is the product of the exponents: (x2)3=x2⋅3=x6.(𝑥2)3=𝑥2⋅3=𝑥6. 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.

(am)n=am⋅n(𝑎𝑚)𝑛=𝑎𝑚⋅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.

53⋅54×5⋅x2(3a)7⋅(3a)10===Product Rule53+4×5+2(3a)7+10===57×7(3a)17butbutbut(53)4(x5)2((3a)7)10Power Rule===53⋅4×5⋅2(3a)7⋅10===512×10(3a)70Product RulePower Rule53⋅54=53+4=57but(53)4=53⋅4=512𝑥5⋅𝑥2=𝑥5+2=𝑥7but(𝑥5)2=𝑥5⋅2=𝑥10(3𝑎)7⋅(3𝑎)10=(3𝑎)7+10=(3𝑎)17but((3𝑎)7)10=(3𝑎)7⋅10=(3𝑎)70

THE POWER RULE OF EXPONENTS

For any real number a𝑎 and positive integers m𝑚 and n,𝑛, the power rule of exponents states that

(am)n=am⋅n(𝑎𝑚)𝑛=𝑎𝑚⋅𝑛

EXAMPLE 3

Using the Power Rule

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

1. ⓐ (x2)7(𝑥2)7
2. ⓑ ((2t)5)3((2𝑡)5)3
3. ⓒ ((−3)5)11((−3)5)11

Solution

Use the power rule to simplify each expression.

1. ⓐ (x2)7=x2⋅7=x14(𝑥2)7=𝑥2⋅7=𝑥14
2. ⓑ ((2t)5)3=(2t)5⋅3=(2t)15((2𝑡)5)3=(2𝑡)5⋅3=(2𝑡)15
3. ⓒ ((−3)5)11=(−3)5⋅11=(−3)55((−3)5)11=(−3)5⋅11=(−3)55

TRY IT #3

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

1. ⓐ ((3y)8)3((3𝑦)8)3
2. ⓑ (t5)7(𝑡5)7
3. ⓒ ((−g)4)4((−𝑔)4)4

Using the Zero Exponent Rule of Exponents

Return to the quotient rule. We made the condition that m>n𝑚>𝑛 so that the difference m−n𝑚−𝑛 would never be zero or negative. What would happen if m=n?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.

t8t8=t8t8=1𝑡8𝑡8=𝑡8𝑡8=1

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

t8t8=t8−8=t0𝑡8𝑡8=𝑡8−8=𝑡0

If we equate the two answers, the result is t0=1.𝑡0=1. This is true for any nonzero real number, or any variable representing a real number.

a0=1𝑎0=1

The sole exception is the expression 00.00. This appears later in more advanced courses, but for now, we will consider the value to be undefined.

THE ZERO EXPONENT RULE OF EXPONENTS

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

a0=1𝑎0=1

EXAMPLE 4

Using the Zero Exponent Rule

Simplify each expression using the zero exponent rule of exponents.

1. ⓐ c3c3𝑐3𝑐3
2. ⓑ −3x5x5−3𝑥5𝑥5
3. ⓒ (j2k)4(j2k)⋅(j2k)3(𝑗2𝑘)4(𝑗2𝑘)⋅(𝑗2𝑘)3
4. ⓓ 5(rs2)2(rs2)25(𝑟𝑠2)2(𝑟𝑠2)2

Solution

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

ⓐ
c3c3===c3−3c01𝑐3𝑐3=𝑐3−3=𝑐0=1

ⓑ
−3x5x5=====−3⋅x5x5−3⋅x5−5−3⋅x0−3⋅1−3−3𝑥5𝑥5=−3⋅𝑥5𝑥5=−3⋅𝑥5−5=−3⋅𝑥0=−3⋅1=−3

ⓒ
(j2k)4(j2k)⋅(j2k)3=====(j2k)4(j2k)1+3(j2k)4(j2k)4(j2k)4−4(j2k)01Use the product rule in the denominator.Simplify.Use the quotient rule.Simplify.(𝑗2𝑘)4(𝑗2𝑘)⋅(𝑗2𝑘)3=(𝑗2𝑘)4(𝑗2𝑘)1+3Use the product rule in the denominator.=(𝑗2𝑘)4(𝑗2𝑘)4Simplify.=(𝑗2𝑘)4−4Use the quotient rule.=(𝑗2𝑘)0Simplify.=1

ⓓ
5(rs2)2(rs2)2====5(rs2)2−25(rs2)05⋅15Use the quotient rule.Simplify.Use the zero exponent rule.Simplify.5(𝑟𝑠2)2(𝑟𝑠2)2=5(𝑟𝑠2)2−2Use the quotient rule.=5(𝑟𝑠2)0Simplify.=5⋅1Use the zero exponent rule.=5Simplify.

TRY IT #4

Simplify each expression using the zero exponent rule of exponents.

1. ⓐ t7t7𝑡7𝑡7
2. ⓑ (de2)112(de2)11(𝑑𝑒2)112(𝑑𝑒2)11
3. ⓒ w4⋅w2w6𝑤4⋅𝑤2𝑤6
4. ⓓ t3⋅t4t2⋅t5𝑡3⋅𝑡4𝑡2⋅𝑡5

Using the Negative Rule of Exponents

Another useful result occurs if we relax the condition that m>n𝑚>𝑛 in the quotient rule even further. For example, can we simplify h3h5?ℎ3ℎ5? When m<n𝑚<𝑛 —that is, where the difference m−n𝑚−𝑛 is negative—we can use the negative rule of exponents to simplify the expression to its reciprocal.

Divide one exponential expression by another with a larger exponent. Use our example, h3h5.ℎ3ℎ5.

h3h5====h⋅h⋅hh⋅h⋅h⋅h⋅hh⋅h⋅hh⋅h⋅h⋅h⋅h1h⋅h1h2ℎ3ℎ5=ℎ⋅ℎ⋅ℎℎ⋅ℎ⋅ℎ⋅ℎ⋅ℎ=ℎ⋅ℎ⋅ℎℎ⋅ℎ⋅ℎ⋅ℎ⋅ℎ=1ℎ⋅ℎ=1ℎ2

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

h3h5==h3−5h−2ℎ3ℎ5=ℎ3−5=ℎ−2

Putting the answers together, we have h−2=1h2.ℎ−2=1ℎ2. 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.

a−n=1anandan=1a−n𝑎−𝑛=1𝑎𝑛and𝑎𝑛=1𝑎−𝑛

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

THE NEGATIVE RULE OF EXPONENTS

For any nonzero real number a𝑎 and natural number n,𝑛, the negative rule of exponents states that

a−n=1an𝑎−𝑛=1𝑎𝑛

EXAMPLE 5

Using the Negative Exponent Rule

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

1. ⓐ θ3θ10𝜃3𝜃10
2. ⓑ z2⋅zz4𝑧2⋅𝑧𝑧4
3. ⓒ (−5t3)4(−5t3)8(−5𝑡3)4(−5𝑡3)8

Solution

1. ⓐ θ3θ10=θ3−10=θ−7=1θ7𝜃3𝜃10=𝜃3−10=𝜃−7=1𝜃7
2. ⓑ z2⋅zz4=z2+1z4=z3z4=z3−4=z−1=1z𝑧2⋅𝑧𝑧4=𝑧2+1𝑧4=𝑧3𝑧4=𝑧3−4=𝑧−1=1𝑧
3. ⓒ (−5t3)4(−5t3)8=(−5t3)4−8=(−5t3)−4=1(−5t3)4(−5𝑡3)4(−5𝑡3)8=(−5𝑡3)4−8=(−5𝑡3)−4=1(−5𝑡3)4

TRY IT #5

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

1. ⓐ (−3t)2(−3t)8(−3𝑡)2(−3𝑡)8
2. ⓑ f47f49⋅f𝑓47𝑓49⋅𝑓
3. ⓒ 2k45k72𝑘45𝑘7

EXAMPLE 6

Using the Product and Quotient Rules

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

1. ⓐ b2⋅b−8𝑏2⋅𝑏−8
2. ⓑ (−x)5⋅(−x)−5(−𝑥)5⋅(−𝑥)−5
3. ⓒ −7z(−7z)5−7𝑧(−7𝑧)5

Solution

1. ⓐ b2⋅b−8=b2−8=b−6=1b6𝑏2⋅𝑏−8=𝑏2−8=𝑏−6=1𝑏6
2. ⓑ (−x)5⋅(−x)−5=(−x)5−5=(−x)0=1(−𝑥)5⋅(−𝑥)−5=(−𝑥)5−5=(−𝑥)0=1
3. ⓒ −7z(−7z)5=(−7z)1(−7z)5=(−7z)1−5=(−7z)−4=1(−7z)4−7𝑧(−7𝑧)5=(−7𝑧)1(−7𝑧)5=(−7𝑧)1−5=(−7𝑧)−4=1(−7𝑧)4

TRY IT #6

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

1. ⓐ t−11⋅t6𝑡−11⋅𝑡6
2. ⓑ 2512251325122513

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.(𝑝𝑞)3. We begin by using the associative and commutative properties of multiplication to regroup the factors.

(pq)3====(pq)⋅(pq)⋅(pq)3factorsp⋅q⋅p⋅q⋅p⋅qp⋅p⋅p3factors⋅q⋅q⋅q3factorsp3⋅q3(𝑝𝑞)3=(𝑝𝑞)⋅(𝑝𝑞)⋅(𝑝𝑞)3factors=𝑝⋅𝑞⋅𝑝⋅𝑞⋅𝑝⋅𝑞=𝑝⋅𝑝⋅𝑝3factors⋅𝑞⋅𝑞⋅𝑞3factors=𝑝3⋅𝑞3

In other words, (pq)3=p3⋅q3.(𝑝𝑞)3=𝑝3⋅𝑞3.

THE POWER OF A PRODUCT RULE OF EXPONENTS

For any real numbers a𝑎 and b𝑏 and any integer n,𝑛, the power of a product rule of exponents states that

(ab)n=anbn(𝑎𝑏)𝑛=𝑎𝑛𝑏𝑛

EXAMPLE 7

Using the Power of a Product Rule

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

1. ⓐ (ab2)3(𝑎𝑏2)3
2. ⓑ (2t)15(2𝑡)15
3. ⓒ (−2w3)3(−2𝑤3)3
4. ⓓ 1(−7z)41(−7𝑧)4
5. ⓔ (e−2f2)7(𝑒−2𝑓2)7

Solution

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

1. ⓐ (ab2)3=(a)3⋅(b2)3=a1⋅3⋅b2⋅3=a3b6(𝑎𝑏2)3=(𝑎)3⋅(𝑏2)3=𝑎1⋅3⋅𝑏2⋅3=𝑎3𝑏6
2. ⓑ (2t)15=(2)15⋅(t)15=215t15=32,768t15(2𝑡)15=(2)15⋅(𝑡)15=215𝑡15=32,768𝑡15
3. ⓒ (−2w3)3=(−2)3⋅(w3)3=−8⋅w3⋅3=−8w9(−2𝑤3)3=(−2)3⋅(𝑤3)3=−8⋅𝑤3⋅3=−8𝑤9
4. ⓓ 1(−7z)4=1(−7)4⋅(z)4=12,401z41(−7𝑧)4=1(−7)4⋅(𝑧)4=12,401𝑧4
5. ⓔ (e−2f2)7=(e−2)7⋅(f2)7=e−2⋅7⋅f2⋅7=e−14f14=f14e14(𝑒−2𝑓2)7=(𝑒−2)7⋅(𝑓2)7=𝑒−2⋅7⋅𝑓2⋅7=𝑒−14𝑓14=𝑓14𝑒14

TRY IT #7

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

1. ⓐ (g2h3)5(𝑔2ℎ3)5
2. ⓑ (5t)3(5𝑡)3
3. ⓒ (−3y5)3(−3𝑦5)3
4. ⓓ 1(a6b7)31(𝑎6𝑏7)3
5. ⓔ (r3s−2)4(𝑟3𝑠−2)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 the following example.

(e−2f2)7=f14e14(𝑒−2𝑓2)7=𝑓14𝑒14

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

(e−2f2)7==(f2e2)7f14e14(𝑒−2𝑓2)7=(𝑓2𝑒2)7=𝑓14𝑒14

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

(e−2f2)7====(f2e2)7(f2)7(e2)7f2⋅7e2⋅7f14e14(𝑒−2𝑓2)7=(𝑓2𝑒2)7=(𝑓2)7(𝑒2)7=𝑓2⋅7𝑒2⋅7=𝑓14𝑒14

THE POWER OF A QUOTIENT RULE OF EXPONENTS

For any real numbers a𝑎 and b𝑏 and any integer n,𝑛, the power of a quotient rule of exponents states that

(ab)n=anbn(𝑎𝑏)𝑛=𝑎𝑛𝑏𝑛

EXAMPLE 8

Using the Power of a Quotient Rule

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

1. ⓐ (4z11)3(4𝑧11)3
2. ⓑ (pq3)6(𝑝𝑞3)6
3. ⓒ (−1t2)27(−1𝑡2)27
4. ⓓ (j3k−2)4(𝑗3𝑘−2)4
5. ⓔ (m−2n−2)3(𝑚−2𝑛−2)3

Solution

1. ⓐ (4z11)3=(4)3(z11)3=64z11⋅3=64z33(4𝑧11)3=(4)3(𝑧11)3=64𝑧11⋅3=64𝑧33
2. ⓑ (pq3)6=(p)6(q3)6=p1⋅6q3⋅6=p6q18(𝑝𝑞3)6=(𝑝)6(𝑞3)6=𝑝1⋅6𝑞3⋅6=𝑝6𝑞18
3. ⓒ (−1t2)27=(−1)27(t2)27=−1t2⋅27=−1t54=−1t54(−1𝑡2)27=(−1)27(𝑡2)27=−1𝑡2⋅27=−1𝑡54=−1𝑡54
4. ⓓ (j3k−2)4=(j3k2)4=(j3)4(k2)4=j3⋅4k2⋅4=j12k8(𝑗3𝑘−2)4=(𝑗3𝑘2)4=(𝑗3)4(𝑘2)4=𝑗3⋅4𝑘2⋅4=𝑗12𝑘8
5. ⓔ (m−2n−2)3=(1m2n2)3=(1)3(m2n2)3=1(m2)3(n2)3=1m2⋅3⋅n2⋅3=1m6n6(𝑚−2𝑛−2)3=(1𝑚2𝑛2)3=(1)3(𝑚2𝑛2)3=1(𝑚2)3(𝑛2)3=1𝑚2⋅3⋅𝑛2⋅3=1𝑚6𝑛6

TRY IT #8

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

1. ⓐ (b5c)3(𝑏5𝑐)3
2. ⓑ (5u8)4(5𝑢8)4
3. ⓒ (−1w3)35(−1𝑤3)35
4. ⓓ (p−4q3)8(𝑝−4𝑞3)8
5. ⓔ (c−5d−3)4(𝑐−5𝑑−3)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.

EXAMPLE 9

Simplifying Exponential Expressions

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

1. ⓐ (6m2n−1)3(6𝑚2𝑛−1)3
2. ⓑ 175⋅17−4⋅17−3175⋅17−4⋅17−3
3. ⓒ (u−1vv−1)2(𝑢−1𝑣𝑣−1)2
4. ⓓ (−2a3b−1)(5a−2b2)(−2𝑎3𝑏−1)(5𝑎−2𝑏2)
5. ⓔ (x22–√)4(x22–√)−4(𝑥22)4(𝑥22)−4
6. ⓕ (3w2)5(6w−2)2(3𝑤2)5(6𝑤−2)2

Solution

1. ⓐ
   (6m2n−1)3====(6)3(m2)3(n−1)363m2⋅3n−1⋅3216m6n−3216m6n3The power of a product ruleThe power ruleSimplify.The negative exponent rule(6𝑚2𝑛−1)3=(6)3(𝑚2)3(𝑛−1)3The power of a product rule=63𝑚2⋅3𝑛−1⋅3The power rule=216𝑚6𝑛−3Simplify.=216𝑚6𝑛3The negative exponent rule
2. ⓑ
   175⋅17−4⋅17−3===175−4−317−21172or 1289The product ruleSimplify.The negative exponent rule175⋅17−4⋅17−3=175−4−3The product rule=17−2Simplify.=1172or 1289The negative exponent rule
3. ⓒ
   (u−1vv−1)2=====(u−1v)2(v−1)2u−2v2v−2u−2v2−(−2)u−2v4v4u2The power of a quotient ruleThe power of a product ruleThe quotient ruleSimplify.The negative exponent rule(𝑢−1𝑣𝑣−1)2=(𝑢−1𝑣)2(𝑣−1)2The power of a quotient rule=𝑢−2𝑣2𝑣−2The power of a product rule=𝑢−2𝑣2−(−2)The quotient rule=𝑢−2𝑣4Simplify.=𝑣4𝑢2The negative exponent rule
4. ⓓ
   (−2a3b−1)(5a−2b2)===−2⋅5⋅a3⋅a−2⋅b−1⋅b2−10⋅a3−2⋅b−1+2−10abCommutative and associative laws of multiplicationThe product ruleSimplify.(−2𝑎3𝑏−1)(5𝑎−2𝑏2)=−2⋅5⋅𝑎3⋅𝑎−2⋅𝑏−1⋅𝑏2Commutative and associative laws of multiplication=−10⋅𝑎3−2⋅𝑏−1+2The product rule=−10𝑎𝑏Simplify.
5. ⓔ
   (x22–√)4(x22–√)−4===(x22–√)4−4(x22–√)01The product ruleSimplify.The zero exponent rule(𝑥22)4(𝑥22)−4=(𝑥22)4−4The product rule=(𝑥22)0Simplify.=1The zero exponent rule
6. ⓕ
   (3w2)5(6w−2)2=====(3)5⋅(w2)5(6)2⋅(w−2)235w2⋅562w−2⋅2243w1036w−427w10−(−4)427w144The power of a product ruleThe power ruleSimplify.The quotient rule and reduce fractionSimplify.(3𝑤2)5(6𝑤−2)2=(3)5⋅(𝑤2)5(6)2⋅(𝑤−2)2The power of a product rule=35𝑤2⋅562𝑤−2⋅2The power rule=243𝑤1036𝑤−4Simplify.=27𝑤10−(−4)4The quotient rule and reduce fraction=27𝑤144Simplify.

TRY IT #9

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

1. ⓐ (2uv−2)−3(2𝑢𝑣−2)−3
2. ⓑ x8⋅x−12⋅x𝑥8⋅𝑥−12⋅𝑥
3. ⓒ (e2f−3f−1)2(𝑒2𝑓−3𝑓−1)2
4. ⓓ (9r−5s3)(3r6s−4)(9𝑟−5𝑠3)(3𝑟6𝑠−4)
5. ⓔ (49tw−2)−3(49tw−2)3(49𝑡𝑤−2)−3(49𝑡𝑤−2)3
6. ⓕ (2h2k)4(7h−1k2)2(2ℎ2𝑘)4(7ℎ−1𝑘2)2

Using Scientific Notation

Recall at the beginning of the section that we found the number 1.3×10131.3×1013 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, n𝑛 is positive. If you moved the decimal right as in a very small number, n𝑛 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.

2.780418×1062.780418×106

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.

4.7×10−134.7×10−13

SCIENTIFIC NOTATION

A number is written in scientific notation if it is written in the form a×10n,𝑎×10𝑛, where 1≤|a|<101≤|𝑎|<10 and n𝑛 is an integer.

EXAMPLE 10

Converting Standard Notation to Scientific Notation

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

Solution

ⓐ
24,000,000,000,000,000,000,000m24,000,000,000,000,000,000,000m←22places2.4×1022m24,000,000,000,000,000,000,000m24,000,000,000,000,000,000,000m←22places2.4×1022m

ⓑ
1,300,000,000,000,000,000,000m1,300,000,000,000,000,000,000m←21places1.3×1021m1,300,000,000,000,000,000,000m1,300,000,000,000,000,000,000m←21places1.3×1021m

ⓒ
1,000,000,000,0001,000,000,000,000←12places1×10121,000,000,000,0001,000,000,000,000←12places1×1012

ⓓ
0.00000000000094m0.00000000000094m→13places9.4×10−13m0.00000000000094m0.00000000000094m→13places9.4×10−13m

ⓔ
0.000001430.00000143→6places1.43×10−60.000001430.00000143→6places1.43×10−6

Analysis

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

TRY IT #10

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

Converting from Scientific to Standard Notation

To convert a number in scientific notation to standard notation, simply reverse the process. Move the decimal n𝑛 places to the right if n𝑛 is positive or n𝑛 places to the left if n𝑛 is negative and add zeros as needed. Remember, if n𝑛 is positive, the absolute value of the number is greater than 1, and if n𝑛 is negative, the absolute value of the number is less than one.

EXAMPLE 11

Converting Scientific Notation to Standard Notation

Convert each number in scientific notation to standard notation.

1. ⓐ 3.547×10143.547×1014
2. ⓑ −2×106−2×106
3. ⓒ 7.91×10−77.91×10−7
4. ⓓ −8.05×10−12−8.05×10−12

Solution

1. ⓐ
   3.547×10143.54700000000000→14places354,700,000,000,0003.547×10143.54700000000000→14places354,700,000,000,000
2. ⓑ
   −2×106−2.000000→6places−2,000,000−2×106−2.000000→6places−2,000,000
3. ⓒ
   7.91×10−70000007.91←7places0.0000007917.91×10−70000007.91←7places0.000000791
4. ⓓ
   −8.05×10−12−000000000008.05←12places−0.00000000000805−8.05×10−12−000000000008.05←12places−0.00000000000805

TRY IT #11

Convert each number in scientific notation to standard notation.

1. ⓐ 7.03×1057.03×105
2. ⓑ −8.16×1011−8.16×1011
3. ⓒ −3.9×10−13−3.9×10−13
4. ⓓ 8×10−68×10−6

Using Scientific Notation in Applications

Scientific notation, used with the rules of exponents, makes calculating with large or small numbers much easier than doing so using standard notation. For example, suppose we are asked to calculate the number of atoms in 1 L of water. Each water molecule contains 3 atoms (2 hydrogen and 1 oxygen). The average drop of water contains around 1.32×10211.32×1021 molecules of water and 1 L of water holds about 1.22×1041.22×104 average drops. Therefore, there are approximately 3⋅(1.32×1021)⋅(1.22×104)≈4.83×10253⋅(1.32×1021)⋅(1.22×104)≈4.83×1025 atoms in 1 L of water. We simply multiply the decimal terms and add the exponents. Imagine having to perform the calculation without using scientific notation!

When performing calculations with scientific notation, be sure to write the answer in proper scientific notation. For example, consider the product (7×104)⋅(5×106)=35×1010.(7×104)⋅(5×106)=35×1010. The answer is not in proper scientific notation because 35 is greater than 10. Consider 35 as 3.5×10.3.5×10. That adds a ten to the exponent of the answer.

(35)×1010=(3.5×10)×1010=3.5×(10×1010)=3.5×1011(35)×1010=(3.5×10)×1010=3.5×(10×1010)=3.5×1011

EXAMPLE 12

Using Scientific Notation

Perform the operations and write the answer in scientific notation.

1. ⓐ (8.14×10−7)(6.5×1010)(8.14×10−7)(6.5×1010)
2. ⓑ (4×105)÷(−1.52×109)(4×105)÷(−1.52×109)
3. ⓒ (2.7×105)(6.04×1013)(2.7×105)(6.04×1013)
4. ⓓ (1.2×108)÷(9.6×105)(1.2×108)÷(9.6×105)
5. ⓔ (3.33×104)(−1.05×107)(5.62×105)(3.33×104)(−1.05×107)(5.62×105)

Solution

1. ⓐ
   (8.14×10−7)(6.5×1010)===(8.14×6.5)(10−7×1010)(52.91)(103)5.291×104Commutative and associativeproperties of multiplicationProduct rule of exponentsScientific notation(8.14×10−7)(6.5×1010)=(8.14×6.5)(10−7×1010)Commutative and associativeproperties of multiplication=(52.91)(103)Product rule of exponents=5.291×104Scientific notation
2. ⓑ
   (4×105)÷(−1.52×109)=≈=(4−1.52)(105109)(−2.63)(10−4)−2.63×10−4Commutative and associativeproperties of multiplicationQuotient rule of exponentsScientific notation(4×105)÷(−1.52×109)=(4−1.52)(105109)Commutative and associativeproperties of multiplication≈(−2.63)(10−4)Quotient rule of exponents=−2.63×10−4Scientific notation
3. ⓒ
   (2.7×105)(6.04×1013)===(2.7×6.04)(105×1013)(16.308)(1018)1.6308×1019Commutative and associativeproperties of multiplicationProduct rule of exponentsScientific notation(2.7×105)(6.04×1013)=(2.7×6.04)(105×1013)Commutative and associativeproperties of multiplication=(16.308)(1018)Product rule of exponents=1.6308×1019Scientific notation
4. ⓓ
   (1.2×108)÷(9.6×105)===(1.29.6)(108105)(0.125)(103)1.25×102Commutative and associativeproperties of multiplicationQuotient rule of exponentsScientific notation(1.2×108)÷(9.6×105)=(1.29.6)(108105)Commutative and associativeproperties of multiplication=(0.125)(103)Quotient rule of exponents=1.25×102Scientific notation
5. ⓔ
   (3.33×104)(−1.05×107)(5.62×105)=≈=[3.33×(−1.05)×5.62](104×107×105)(−19.65)(1016)−1.965×1017(3.33×104)(−1.05×107)(5.62×105)=[3.33×(−1.05)×5.62](104×107×105)≈(−19.65)(1016)=−1.965×1017

TRY IT #12

Perform the operations and write the answer in scientific notation.

1. ⓐ (−7.5×108)(1.13×10−2)(−7.5×108)(1.13×10−2)
2. ⓑ (1.24×1011)÷(1.55×1018)(1.24×1011)÷(1.55×1018)
3. ⓒ (3.72×109)(8×103)(3.72×109)(8×103)
4. ⓓ (9.933×1023)÷(−2.31×1017)(9.933×1023)÷(−2.31×1017)
5. ⓔ (−6.04×109)(7.3×102)(−2.81×102)(−6.04×109)(7.3×102)(−2.81×102)

EXAMPLE 13

Applying Scientific Notation to Solve Problems

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.

Solution

The population was 308,000,000=3.08×108.308,000,000=3.08×108.

The national debt was $17,547,000,000,000≈$1.75×1013.$17,547,000,000,000≈$1.75×1013.

To find the amount of debt per citizen, divide the national debt by the number of citizens.

(1.75×1013)÷(3.08×108)=≈=(1.753.08)⋅(1013108)0.57×1055.7×104(1.75×1013)÷(3.08×108)=(1.753.08)⋅(1013108)≈0.57×105=5.7×104

The debt per citizen at the time was about $5.7×104,$5.7×104, or $57,000.

TRY IT #13

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.