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Study Guides > College Algebra

Exponents and 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 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 2,048×1,536×48×24×3,6002,048\times 1,536\times 48\times 24\times 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\times {10}^{13} 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.

Rules for Exponents

Consider the product x3⋅x4{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}\text{ }x^{3}\cdot x^{4}\hfill&=\stackrel{\text{3 factors } \text{ 4 factors}}{x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x} \\ \hfill& =\stackrel{7 \text{ factors}}{x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x} \\ \hfill& =x^{7}\end{array}
The result is that x3⋅x4=x3+4=x7{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 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+n{a}^{m}\cdot {a}^{n}={a}^{m+n}
Now consider an example with real numbers.
23⋅24=23+4=27{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 23{2}^{3} is 8, 24{2}^{4} is 16, and 27{2}^{7} is 128. The product 8⋅168\cdot 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.

A General Note: The Product Rule of Exponents

For any real number aa and natural numbers mm and nn, the product rule of exponents states that
am⋅an=am+n{a}^{m}\cdot {a}^{n}={a}^{m+n}

Example: Using the Product Rule

Write each of the following products with a single base. Do not simplify further.
  1. t5⋅t3{t}^{5}\cdot {t}^{3}
  2. (−3)5⋅(−3)\left(-3\right)^{5}\cdot \left(-3\right)
  3. x2⋅x5⋅x3{x}^{2}\cdot {x}^{5}\cdot {x}^{3}

Answer: Use the product rule to simplify each expression.

  1. t5⋅t3=t5+3=t8{t}^{5}\cdot {t}^{3}={t}^{5+3}={t}^{8}
  2. (−3)5⋅(−3)=(−3)5⋅(−3)1=(−3)5+1=(−3)6{\left(-3\right)}^{5}\cdot \left(-3\right)={\left(-3\right)}^{5}\cdot {\left(-3\right)}^{1}={\left(-3\right)}^{5+1}={\left(-3\right)}^{6}
  3. x2⋅x5⋅x3{x}^{2}\cdot {x}^{5}\cdot {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.
x2⋅x5⋅x3=(x2⋅x5)⋅x3=(x2+5)⋅x3=x7⋅x3=x7+3=x10{x}^{2}\cdot {x}^{5}\cdot {x}^{3}=\left({x}^{2}\cdot {x}^{5}\right)\cdot {x}^{3}=\left({x}^{2+5}\right)\cdot {x}^{3}={x}^{7}\cdot {x}^{3}={x}^{7+3}={x}^{10}
Notice we get the same result by adding the three exponents in one step.
x2⋅x5⋅x3=x2+5+3=x10{x}^{2}\cdot {x}^{5}\cdot {x}^{3}={x}^{2+5+3}={x}^{10}

Try It

Write each of the following products with a single base. Do not simplify further.
  1. k6⋅k9{k}^{6}\cdot {k}^{9}
  2. (2y)4⋅(2y){\left(\frac{2}{y}\right)}^{4}\cdot \left(\frac{2}{y}\right)
  3. t3⋅t6⋅t5{t}^{3}\cdot {t}^{6}\cdot {t}^{5}

Answer:

  1. k15{k}^{15}
  2. (2y)5{\left(\frac{2}{y}\right)}^{5}
  3. t14{t}^{14}

Zero and Negative 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=nm=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=1\frac{t^{8}}{t^{8}}=\frac{\cancel{t^{8}}}{\cancel{t^{8}}}=1

If we were to simplify the original expression using the quotient rule, we would have
t8t8=t8−8=t0\frac{{t}^{8}}{{t}^{8}}={t}^{8 - 8}={t}^{0}
If we equate the two answers, the result is t0=1{t}^{0}=1. This is true for any nonzero real number, or any variable representing a real number.
a0=1{a}^{0}=1
The sole exception is the expression 00{0}^{0}. This appears later in more advanced courses, but for now, we will consider the value to be undefined.

A General Note: The Zero Exponent Rule of Exponents

For any nonzero real number aa, the zero exponent rule of exponents states that
a0=1{a}^{0}=1

Example: Using the Zero Exponent Rule

Simplify each expression using the zero exponent rule of exponents.
  1. c3c3\frac{{c}^{3}}{{c}^{3}}
  2. −3x5x5\frac{-3{x}^{5}}{{x}^{5}}
  3. (j2k)4(j2k)⋅(j2k)3\frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}
  4. 5(rs2)2(rs2)2\frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}

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

  1. \begin{array}\text{ }\frac{c^{3}}{c^{3}} \hfill& =c^{3-3} \\ \hfill& =c^{0} \\ \hfill& =1\end{array}
  2. −3x5x5=−3⋅x5x5=−3⋅x5−5=−3⋅x0=−3⋅1=−3\begin{array}{ccc}\hfill \frac{-3{x}^{5}}{{x}^{5}}& =& -3\cdot \frac{{x}^{5}}{{x}^{5}}\hfill \\ & =& -3\cdot {x}^{5 - 5}\hfill \\ & =& -3\cdot {x}^{0}\hfill \\ & =& -3\cdot 1\hfill \\ & =& -3\hfill \end{array}
  3. (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\begin{array}{cccc}\hfill \frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}& =& \frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{1+3}}\hfill & \text{Use the product rule in the denominator}.\hfill \\ & =& \frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{4}}\hfill & \text{Simplify}.\hfill \\ & =& {\left({j}^{2}k\right)}^{4 - 4}\hfill & \text{Use the quotient rule}.\hfill \\ & =& {\left({j}^{2}k\right)}^{0}\hfill & \text{Simplify}.\hfill \\ & =& 1& \end{array}
  4. 5(rs2)2(rs2)2=5(rs2)2−2Use the quotient rule.=5(rs2)0Simplify.=5⋅1Use the zero exponent rule.=5Simplify.\begin{array}{cccc}\hfill \frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}& =& 5{\left(r{s}^{2}\right)}^{2 - 2}\hfill & \text{Use the quotient rule}.\hfill \\ & =& 5{\left(r{s}^{2}\right)}^{0}\hfill & \text{Simplify}.\hfill \\ & =& 5\cdot 1\hfill & \text{Use the zero exponent rule}.\hfill \\ & =& 5\hfill & \text{Simplify}.\hfill \end{array}

Try It

Simplify each expression using the zero exponent rule of exponents.
  1. t7t7\frac{{t}^{7}}{{t}^{7}}
  2. (de2)112(de2)11\frac{{\left(d{e}^{2}\right)}^{11}}{2{\left(d{e}^{2}\right)}^{11}}
  3. w4⋅w2w6\frac{{w}^{4}\cdot {w}^{2}}{{w}^{6}}
  4. t3⋅t4t2⋅t5\frac{{t}^{3}\cdot {t}^{4}}{{t}^{2}\cdot {t}^{5}}

Answer:

  1. 11
  2. 12\frac{1}{2}
  3. 11
  4. 11

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\frac{{h}^{3}}{{h}^{5}}? 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\frac{{h}^{3}}{{h}^{5}}.
h3h5=h⋅h⋅hh⋅h⋅h⋅h⋅h=h⋅h⋅hh⋅h⋅h⋅h⋅h=1h⋅h=1h2\begin{array}{ccc}\hfill \frac{{h}^{3}}{{h}^{5}}& =& \frac{h\cdot h\cdot h}{h\cdot h\cdot h\cdot h\cdot h}\hfill \\ & =& \frac{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}}{\cancel{h}\cdot \cancel{h}\cdot \cancel{h}\cdot h\cdot h}\hfill \\ & =& \frac{1}{h\cdot h}\hfill \\ & =& \frac{1}{{h}^{2}}\hfill \end{array}
If we were to simplify the original expression using the quotient rule, we would have
h3h5=h3−5= h−2\begin{array}{ccc}\hfill \frac{{h}^{3}}{{h}^{5}}& =& {h}^{3 - 5}\hfill \\ & =& \text{ }{h}^{-2}\hfill \end{array}
Putting the answers together, we have h−2=1h2{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.
a−n=1anandan=1a−n\begin{array}{ccc}{a}^{-n}=\frac{1}{{a}^{n}}& \text{and}& {a}^{n}=\frac{1}{{a}^{-n}}\end{array}
We have shown that the exponential expression an{a}^{n} is defined when nn is a natural number, 0, or the negative of a natural number. That means that an{a}^{n} is defined for any integer nn. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer nn.

A General Note: The Negative Rule of Exponents

For any nonzero real number aa and natural number nn, the negative rule of exponents states that
a−n=1an{a}^{-n}=\frac{1}{{a}^{n}}

Example: 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. θ3θ10\frac{{\theta }^{3}}{{\theta }^{10}}
  2. z2⋅zz4\frac{{z}^{2}\cdot z}{{z}^{4}}
  3. (−5t3)4(−5t3)8\frac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}

Answer:

  1. θ3θ10=θ3−10=θ−7=1θ7\frac{{\theta }^{3}}{{\theta }^{10}}={\theta }^{3 - 10}={\theta }^{-7}=\frac{1}{{\theta }^{7}}
  2. z2⋅zz4=z2+1z4=z3z4=z3−4=z−1=1z\frac{{z}^{2}\cdot z}{{z}^{4}}=\frac{{z}^{2+1}}{{z}^{4}}=\frac{{z}^{3}}{{z}^{4}}={z}^{3 - 4}={z}^{-1}=\frac{1}{z}
  3. (−5t3)4(−5t3)8=(−5t3)4−8=(−5t3)−4=1(−5t3)4\frac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}={\left(-5{t}^{3}\right)}^{4 - 8}={\left(-5{t}^{3}\right)}^{-4}=\frac{1}{{\left(-5{t}^{3}\right)}^{4}}

Try It

Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents.
  1. (−3t)2(−3t)8\frac{{\left(-3t\right)}^{2}}{{\left(-3t\right)}^{8}}
  2. f47f49⋅f\frac{{f}^{47}}{{f}^{49}\cdot f}
  3. 2k45k7\frac{2{k}^{4}}{5{k}^{7}}

Answer:

  1. 1(−3t)6\frac{1}{{\left(-3t\right)}^{6}}
  2. 1f3\frac{1}{{f}^{3}}
  3. 25k3\frac{2}{5{k}^{3}}

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{\left(pq\right)}^{3}. We begin by using the associative and commutative properties of multiplication to regroup the factors.
(pq)3=(pq)⋅(pq)⋅(pq)3 factors=p⋅q⋅p⋅q⋅p⋅q=p⋅p⋅p3 factors⋅q⋅q⋅q3 factors=p3⋅q3\begin{array}{ccc}\hfill {\left(pq\right)}^{3}& =& \stackrel{3\text{ factors}}{{\left(pq\right)\cdot \left(pq\right)\cdot \left(pq\right)}}\hfill \\ & =& p\cdot q\cdot p\cdot q\cdot p\cdot q\hfill \\ & =& \stackrel{3\text{ factors}}{{p\cdot p\cdot p}}\cdot \stackrel{3\text{ factors}}{{q\cdot q\cdot q}}\hfill \\ & =& {p}^{3}\cdot {q}^{3}\hfill \end{array}
In other words, (pq)3=p3⋅q3{\left(pq\right)}^{3}={p}^{3}\cdot {q}^{3}.

A General Note: The Power of a Product Rule of Exponents

For any real numbers aa and bb and any integer nn, the power of a product rule of exponents states that
(ab)n=anbn{\left(ab\right)}^{n}={a}^{n}{b}^{n}

Example: 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{\left(a{b}^{2}\right)}^{3}
  2. (2t)15{\left(2t\right)}^{15}
  3. (−2w3)3{\left(-2{w}^{3}\right)}^{3}
  4. 1(−7z)4\frac{1}{{\left(-7z\right)}^{4}}
  5. (e−2f2)7{\left({e}^{-2}{f}^{2}\right)}^{7}

Answer: 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{\left(a{b}^{2}\right)}^{3}={\left(a\right)}^{3}\cdot {\left({b}^{2}\right)}^{3}={a}^{1\cdot 3}\cdot {b}^{2\cdot 3}={a}^{3}{b}^{6}
  2. 2t15=(2)15⋅(t)15=215t15=32,768t152{t}^{15}={\left(2\right)}^{15}\cdot {\left(t\right)}^{15}={2}^{15}{t}^{15}=32,768{t}^{15}
  3. (−2w3)3=(−2)3⋅(w3)3=−8⋅w3⋅3=−8w9{\left(-2{w}^{3}\right)}^{3}={\left(-2\right)}^{3}\cdot {\left({w}^{3}\right)}^{3}=-8\cdot {w}^{3\cdot 3}=-8{w}^{9}
  4. 1(−7z)4=1(−7)4⋅(z)4=12,401z4\frac{1}{{\left(-7z\right)}^{4}}=\frac{1}{{\left(-7\right)}^{4}\cdot {\left(z\right)}^{4}}=\frac{1}{2,401{z}^{4}}
  5. (e−2f2)7=(e−2)7⋅(f2)7=e−2⋅7⋅f2⋅7=e−14f14=f14e14{\left({e}^{-2}{f}^{2}\right)}^{7}={\left({e}^{-2}\right)}^{7}\cdot {\left({f}^{2}\right)}^{7}={e}^{-2\cdot 7}\cdot {f}^{2\cdot 7}={e}^{-14}{f}^{14}=\frac{{f}^{14}}{{e}^{14}}

Try It

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{\left({g}^{2}{h}^{3}\right)}^{5}
  2. (5t)3{\left(5t\right)}^{3}
  3. (−3y5)3{\left(-3{y}^{5}\right)}^{3}
  4. 1(a6b7)3\frac{1}{{\left({a}^{6}{b}^{7}\right)}^{3}}
  5. (r3s−2)4{\left({r}^{3}{s}^{-2}\right)}^{4}

Answer:

  1. g10h15{g}^{10}{h}^{15}
  2. 125t3125{t}^{3}
  3. −27y15-27{y}^{15}
  4. 1a18b21\frac{1}{{a}^{18}{b}^{21}}
  5. r12s8\frac{{r}^{12}}{{s}^{8}}

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{\left({e}^{-2}{f}^{2}\right)}^{7}=\frac{{f}^{14}}{{e}^{14}}
Let’s rewrite the original problem differently and look at the result.
(e−2f2)7=(f2e2)7=f14e14\begin{array}{ccc}\hfill {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\hfill \\ & =& \frac{{f}^{14}}{{e}^{14}}\hfill \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.
(e−2f2)7=(f2e2)7=(f2)7(e2)7=f2⋅7e2⋅7=f14e14\begin{array}{ccc}\hfill {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\hfill \\ & =& \frac{{\left({f}^{2}\right)}^{7}}{{\left({e}^{2}\right)}^{7}}\hfill \\ & =& \frac{{f}^{2\cdot 7}}{{e}^{2\cdot 7}}\hfill \\ & =& \frac{{f}^{14}}{{e}^{14}}\hfill \end{array}

A General Note: The Power of a Quotient Rule of Exponents

For any real numbers aa and bb and any integer nn, the power of a quotient rule of exponents states that
(ab)n=anbn{\left(\frac{a}{b}\right)}^{n}=\frac{{a}^{n}}{{b}^{n}}

Example: 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{\left(\frac{4}{{z}^{11}}\right)}^{3}
  2. (pq3)6{\left(\frac{p}{{q}^{3}}\right)}^{6}
  3. (−1t2)27{\left(\frac{-1}{{t}^{2}}\right)}^{27}
  4. (j3k−2)4{\left({j}^{3}{k}^{-2}\right)}^{4}
  5. (m−2n−2)3{\left({m}^{-2}{n}^{-2}\right)}^{3}

Answer:

  1. (4z11)3=(4)3(z11)3=64z11⋅3=64z33{\left(\frac{4}{{z}^{11}}\right)}^{3}=\frac{{\left(4\right)}^{3}}{{\left({z}^{11}\right)}^{3}}=\frac{64}{{z}^{11\cdot 3}}=\frac{64}{{z}^{33}}
  2. (pq3)6=(p)6(q3)6=p1⋅6q3⋅6=p6q18{\left(\frac{p}{{q}^{3}}\right)}^{6}=\frac{{\left(p\right)}^{6}}{{\left({q}^{3}\right)}^{6}}=\frac{{p}^{1\cdot 6}}{{q}^{3\cdot 6}}=\frac{{p}^{6}}{{q}^{18}}
  3. left(−1t2right)27=left(−1right)27left(t2right)27=−1t2⋅27=−1t54=−1t54{\\left(\frac{-1}{{t}^{2}}\\right)}^{27}=\frac{{\\left(-1\\right)}^{27}}{{\\left({t}^{2}\\right)}^{27}}=\frac{-1}{{t}^{2\cdot 27}}=\frac{-1}{{t}^{54}}=-\frac{1}{{t}^{54}}
  4. (j3k−2)4=(j3k2)4=(j3)4(k2)4=j3⋅4k2⋅4=j12k8{\left({j}^{3}{k}^{-2}\right)}^{4}={\left(\frac{{j}^{3}}{{k}^{2}}\right)}^{4}=\frac{{\left({j}^{3}\right)}^{4}}{{\left({k}^{2}\right)}^{4}}=\frac{{j}^{3\cdot 4}}{{k}^{2\cdot 4}}=\frac{{j}^{12}}{{k}^{8}}
  5. (m−2n−2)3=(1m2n2)3=(1)3(m2n2)3=1(m2)3(n2)3=1m2⋅3⋅n2⋅3=1m6n6{\left({m}^{-2}{n}^{-2}\right)}^{3}={\left(\frac{1}{{m}^{2}{n}^{2}}\right)}^{3}=\frac{{\left(1\right)}^{3}}{{\left({m}^{2}{n}^{2}\right)}^{3}}=\frac{1}{{\left({m}^{2}\right)}^{3}{\left({n}^{2}\right)}^{3}}=\frac{1}{{m}^{2\cdot 3}\cdot {n}^{2\cdot 3}}=\frac{1}{{m}^{6}{n}^{6}}

Try It

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{\left(\frac{{b}^{5}}{c}\right)}^{3}
  2. (5u8)4{\left(\frac{5}{{u}^{8}}\right)}^{4}
  3. (−1w3)35{\left(\frac{-1}{{w}^{3}}\right)}^{35}
  4. (p−4q3)8{\left({p}^{-4}{q}^{3}\right)}^{8}
  5. (c−5d−3)4{\left({c}^{-5}{d}^{-3}\right)}^{4}

Answer:

  1. b15c3\frac{{b}^{15}}{{c}^{3}}
  2. 625u32\frac{625}{{u}^{32}}
  3. −1w105\frac{-1}{{w}^{105}}
  4. q24p32\frac{{q}^{24}}{{p}^{32}}
  5. 1c20d12\frac{1}{{c}^{20}{d}^{12}}

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\times {10}^{21} molecules of water and 1 L of water holds about 1.22×1041.22\times {10}^{4} average drops. Therefore, there are approximately 3⋅(1.32×1021)⋅(1.22×104)≈4.83×10253\cdot \left(1.32\times {10}^{21}\right)\cdot \left(1.22\times {10}^{4}\right)\approx 4.83\times {10}^{25} 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\left(7\times {10}^{4}\right)\cdot \left(5\times {10}^{6}\right)=35\times {10}^{10}. The answer is not in proper scientific notation because 35 is greater than 10. Consider 35 as 3.5×103.5\times 10. That adds a ten to the exponent of the answer.
(35)×1010=(3.5×10)×1010=3.5×(10×1010)=3.5×1011\left(35\right)\times {10}^{10}=\left(3.5\times 10\right)\times {10}^{10}=3.5\times \left(10\times {10}^{10}\right)=3.5\times {10}^{11}

Example: Using Scientific Notation

Perform the operations and write the answer in scientific notation.
  1. (8.14×10−7)(6.5×1010)\left(8.14\times {10}^{-7}\right)\left(6.5\times {10}^{10}\right)
  2. (4×105)÷(−1.52×109)\left(4\times {10}^{5}\right)\div \left(-1.52\times {10}^{9}\right)
  3. (2.7×105)(6.04×1013)\left(2.7\times {10}^{5}\right)\left(6.04\times {10}^{13}\right)
  4. (1.2×108)÷(9.6×105)\left(1.2\times {10}^{8}\right)\div \left(9.6\times {10}^{5}\right)
  5. (3.33×104)(−1.05×107)(5.62×105)\left(3.33\times {10}^{4}\right)\left(-1.05\times {10}^{7}\right)\left(5.62\times {10}^{5}\right)

Answer:

  1. \begin{array}\text{ }\left(8.14 \times 10^{-7}\right)\left(6.5 \times 10^{10}\right) \hfill& =\left(8.14 \times 6.5\right)\left(10^{-7} \times 10^{10}\right) \hfill& \text{Commutative and associative properties of multiplication} \\ \hfill& =\left(52.91\right)\left(10^{3}\right) \hfill& \text{Product rule of exponents} \\ \hfill& =5.291 \times 10^{4} \hfill& \text{Scientific notation}\end{array}
  2. (4×105)÷(−1.52×109)=(4−1.52)(105109)Commutative and associative properties of multiplication=(−2.63)(10−4)Quotient rule of exponents=−2.63×10−4Scientific notation\begin{array}{cccc}\hfill \left(4\times {10}^{5}\right)\div \left(-1.52\times {10}^{9}\right)& =& \left(\frac{4}{-1.52}\right)\left(\frac{{10}^{5}}{{10}^{9}}\right)\hfill & \text{Commutative and associative properties of multiplication}\hfill \\ & =& \left(-2.63\right)\left({10}^{-4}\right)\hfill & \text{Quotient rule of exponents}\hfill \\ & =& -2.63\times {10}^{-4}\hfill & \text{Scientific notation}\hfill \end{array}
  3. (2.7×105)(6.04×1013)=(2.7×6.04)(105×1013)Commutative and associative properties of multiplication=(16.308)(1018)Product rule of exponents=1.6308×1019Scientific notation\begin{array}{cccc}\hfill \left(2.7\times {10}^{5}\right)\left(6.04\times {10}^{13}\right)& =& \left(2.7\times 6.04\right)\left({10}^{5}\times {10}^{13}\right)\hfill & \text{Commutative and associative properties of multiplication}\hfill \\ & =& \left(16.308\right)\left({10}^{18}\right)\hfill & \text{Product rule of exponents}\hfill \\ & =& 1.6308\times {10}^{19}\hfill & \text{Scientific notation}\hfill \end{array}
  4. (1.2×108)÷(9.6×105)=(1.29.6)(108105)Commutative and associative properties of multiplication=(0.125)(103)Quotient rule of exponents=1.25×102Scientific notation\begin{array}{cccc}\hfill \left(1.2\times {10}^{8}\right)\div \left(9.6\times {10}^{5}\right)& =& \left(\frac{1.2}{9.6}\right)\left(\frac{{10}^{8}}{{10}^{5}}\right)\hfill & \text{Commutative and associative properties of multiplication}\hfill \\ & =& \left(0.125\right)\left({10}^{3}\right)\hfill & \text{Quotient rule of exponents}\hfill \\ & =& 1.25\times {10}^{2}\hfill & \text{Scientific notation}\hfill \end{array}
  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\begin{array}{ccc}\hfill \left(3.33\times {10}^{4}\right)\left(-1.05\times {10}^{7}\right)\left(5.62\times {10}^{5}\right)& =& \left[3.33\times \left(-1.05\right)\times 5.62\right]\left({10}^{4}\times {10}^{7}\times {10}^{5}\right)\hfill \\ & \approx & \left(-19.65\right)\left({10}^{16}\right)\hfill \\ & =& -1.965\times {10}^{17}\hfill \end{array}

Try It

Perform the operations and write the answer in scientific notation.
  1. (−7.5×108)(1.13×10−2)\left(-7.5\times {10}^{8}\right)\left(1.13\times {10}^{-2}\right)
  2. (1.24×1011)÷(1.55×1018)\left(1.24\times {10}^{11}\right)\div \left(1.55\times {10}^{18}\right)
  3. (3.72×109)(8×103)\left(3.72\times {10}^{9}\right)\left(8\times {10}^{3}\right)
  4. (9.933×1023)÷(−2.31×1017)\left(9.933\times {10}^{23}\right)\div \left(-2.31\times {10}^{17}\right)
  5. (−6.04×109)(7.3×102)(−2.81×102)\left(-6.04\times {10}^{9}\right)\left(7.3\times {10}^{2}\right)\left(-2.81\times {10}^{2}\right)

Answer:

  1. −8.475×106-8.475\times {10}^{6}
  2. 8×10−88\times {10}^{-8}
  3. 2.976×10132.976\times {10}^{13}
  4. −4.3×106-4.3\times {10}^{6}
  5. ≈1.24×1015\approx 1.24\times {10}^{15}

Key Equations

Rules of Exponents For nonzero real numbers aa and bb and integers mm and nn
Product rule am⋅an=am+n{a}^{m}\cdot {a}^{n}={a}^{m+n}
Quotient rule aman=am−n\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}
Power rule (am)n=am⋅n{\left({a}^{m}\right)}^{n}={a}^{m\cdot n}
Zero exponent rule a0=1{a}^{0}=1
Negative rule a−n=1an{a}^{-n}=\frac{1}{{a}^{n}}
Power of a product rule (a⋅b)n=an⋅bn{\left(a\cdot b\right)}^{n}={a}^{n}\cdot {b}^{n}
Power of a quotient rule (ab)n=anbn{\left(\frac{a}{b}\right)}^{n}=\frac{{a}^{n}}{{b}^{n}}

Key Concepts

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

Glossary

scientific notation a shorthand notation for writing very large or very small numbers in the form a×10na\times {10}^{n} where 1≤∣a∣<101\le |a|<10 and nn is an integer

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