ORCCA (Draft)Open Resources for Community College Algebra

Without knowing a rule for simplifying this quotient of powers, we can write the expressions without exponents and simplify.

Notice that the difference of the exponents of the numerator and the denominator (\(5\) and \(2\text{,}\) respectively) is \(3\text{,}\) which is the exponent of the simplified expression.

Writing the expression without an exponent and then simplifying, we have:

Similar to the product to a power rule, we essentially applied the outer exponent to the entire numerator and denominator.

1. If we stick closely to the order of operations, we should first simplify inside the parentheses and then work with the outer exponent. Going this route, we will first use the quotient rule:

   \begin{align*} \left(\dfrac{5^8w^7}{5^2w^4}\right)^9 \amp= \left(5^{8-2}w^{7-4}\right)^9 \\ \amp= \left(5^{6}w^{3}\right)^9 \\ \end{align*}

   Now we can apply the outer exponent to each factor inside the parentheses using the product to a power rule.

   \begin{align*} \phantom{\left(\dfrac{5^8w^7}{5^2w^4}\right)^9 }\amp= \left(5^{6}\right)^9\cdot \left(w^{3}\right)^9 \\ \end{align*}

   To finish, we need to use the power to a power rule.

   \begin{align*} \phantom{\left(\dfrac{5^8w^7}{5^2w^4}\right)^9 }\amp= 5^{6\cdot 9}\cdot w^{3 \cdot 9} \\ \amp= 5^{54}\cdot w^{27} \end{align*}

2. According to the order of operations, we should simplify inside parentheses first, then apply exponents, then divide. Since we cannot simplify inside the parentheses, we must apply the outer exponents to each factor inside the respective set of parentheses first:

   \begin{align*} \dfrac{\left(2r^5\right)^7}{\left(2^2 r^8\right)^3} \amp= \dfrac{\left(2\right)^7 \left(r^5\right)^7}{\left(2^2\right)^3 \left(r^8\right)^3} \\ \end{align*}

   At this point, we need to use the power-to-a-power rule:

   \begin{align*} \phantom{\dfrac{\left(2r^5\right)^7}{\left(2^2 r^8\right)^3}} \amp= \dfrac{2^{7} r^{5\cdot 7}}{ 2^{2\cdot 3} r^{8 \cdot 3}} \\ \amp= \dfrac{2^{7} r^{35}}{ 2^{6} r^{24}} \\ \end{align*}

   To finish simplifying, we'll conclude with the quotient rule:

   \begin{align*} \phantom{\dfrac{\left(2r^5\right)^7}{\left(2^2 r^8\right)^3}} \amp= 2^{7-6} r^{35-24} \\ \amp= 2^{1} r^{11} \\ \amp= 2 r^{11} \end{align*}

To simplify any of these expressions, it is critical that we remember an exponent only applies to what it is touching or immediately next to.

1. In the expression \(\left(173 x^4 y^{251}\right)^0\text{,}\) the exponent \(0\) applies to everything inside the parentheses.

   \begin{align*} \left(173 x^4 y^{251}\right)^0 \amp= 1 \end{align*}

2. In the expression \((-8)^0\) the exponent applies to everything inside the parentheses, \(-8\text{.}\)

   \begin{align*} (-8)^0 \amp= 1 \end{align*}

3. In contrast to the previous example, the exponent only applies to the \(8\text{.}\) We should consider that \(-8^0 = -1 \cdot 8^0\text{,}\) and so:

   \begin{align*} -8^0 \amp= -1\cdot 8^0 \\ \amp= -1\cdot 1 \\ \amp= -1 \end{align*}

4. In the expression \(3x^0\text{,}\) the exponent \(0\) only applies to the \(x\text{:}\)

   \begin{align*} 3x^0 \amp= 3\cdot x^0 \\ \amp= 3\cdot 1 \\ \amp= 3 \end{align*}

   Since the exponent did not apply to the \(3\text{,}\) the \(3\) stayed a \(3\text{.}\)

1. We will apply the zero exponent rule:

   \begin{align*} 5^{-3} \amp= \frac{1}{5^{3}} \\ \amp= \frac{1}{125} \end{align*}

2. Always remember that an exponent only applies to what it is touching. In the expression \(4y^{-6}\text{,}\) only the \(y\) has an exponent of \(-6\text{.}\)

   \begin{align*} 4y^{-6} \amp= 4\cdot \frac{1}{y^{6}} \\ \amp= \frac{4}{y^6} \end{align*}

3. We will first use the quotient to a power rule, and then use the >negative exponent rule:

   \begin{align*} \left(\frac{3}{7}\right)^{-2} \amp= \left(\frac{3^{-2}}{7^{-2}}\right) \\ \amp= \frac{7^2}{3^2} \\ \amp= \frac{49}{9} \end{align*}

Notice that the factors of \(3\) and \(y\) did not move, as both of those factors had positive exponents.

1. In the expression \(\frac{6x^{3}}{2x^{7}}\text{,}\) the coefficients reduce using the properties of fractions. One way to simplify the variables components is to cancel them:

   \begin{align*} \frac{6x^{3}}{2x^{7}} \amp= \frac{6}{2} \cdot \frac{x^{3}}{x^{7}}\\ \amp= 3 \cdot \frac{1}{x^{4}}\\ \amp= \frac{3}{x^4} \end{align*}

   Alternately, we could have turned \(\frac{x^3}{x^7}\) into \(x^{-4}\) before turning it back into \(\frac{1}{x^4}\text{.}\)

2. In the expression \(4\left(\frac{1}{5}tv^{-4}\right)^{2}\text{,}\) the exponent \(2\) applies to the expression inside the parentheses.

   \begin{align*} 4\left(\frac{1}{5}tv^{-4}\right)^{2} \amp= 4\left(\frac{1}{5}\right)^{2}(t^2)\left(v^{-4}\right)^2\\ \amp= 4\left(\frac{1}{25}\right)(t^2)\left(v^{-4\cdot 2}\right)\\ \amp= 4\left(\frac{1}{25}\right)(t^2)\left(v^{-8}\right)\\ \amp= 4\left(\frac{1}{25}\right)(t^2)\left(\frac{1}{v^{8}}\right)\\ \amp= \frac{4t^2}{25v^{8}} \end{align*}

3. To follow the order of operations in the expression \(\left(\frac{5^{0}y^{4}\cdot y^{5}}{5y^2}\right)^{3}\text{,}\) the numerator inside the parentheses should be dealt with first. After that, we'll simplify the quotient inside the parentheses. As a final step, we'll apply the exponent to that simplified expression:

   \begin{align*} \left(\frac{3^0 y^4 \cdot y^5}{6y^2}\right)^3 \amp= \left(\frac{1\cdot y^{4+5}}{6y^2}\right)^3\\ \amp= \left(\frac{y^{9}}{6y^2}\right)^{3}\\ \amp= \left(\frac{y^7}{6}\right)^3\\ \amp= \frac{\left(y^7\right)^3}{6^3}\\ \amp= \frac{y^{7\cdot 3}}{216}\\ \amp= \frac{y^{21}}{216} \end{align*}

We'll again rely on the order of operations, and look to simplify anything inside parentheses first and then apply exponents. In this example, we will begin by applying the product to a power rule, followed by the power to a power rule.

1. \(5.7 \times 10 = 57\)

   \(10 = 10^1\)and multiplying by \(10^1\) moved the decimal point one place to the right.

2. \(3.1 \times 10000=31{,}000\)

   \(10,000 = 10^4\) and multiplying by \(10^4\) moved the decimal point four places to the right.

1. To convert the federal debt to scientific notation, we will count the number of digits after the first non-zero digit \(1\text{.}\) Since there are 13 places after the first non-zero digit, we write:

   \begin{equation*} 1\overbrace{9{,}600{,}000{,}000{,}000}^{13\text{ places}} \text{ dollars} = 1.96\times 10^{13} \text{ dollars} \end{equation*}

2. Since there are nine places after the first non-zero digit of \(7\text{,}\) the world's population in 2016 was about

   \begin{equation*} 7{,}\overbrace{418{,}000{,}000}^{9\text{ places}} \text{ people} = 7.418 \times 10^{9} \text{ people} \end{equation*}

1. To convert this number to decimal notation we will move the decimal point after the digit \(1\) nine places to the right, including zeros where necessary. The earth's diameter is:

   \begin{equation*} 1.27\times 10^{7}\text{ meters} = 1\overbrace{2{,}700{,}000}^{7\text{ places}}\text{ meters} \end{equation*}

2. Avogadro's number is about:

   \begin{equation*} 6.022\times 10^{23} = 6\overbrace{02{,}200{,}000{,}000{,}000{,}000{,}000{,}000}^{23\text{ places}} \end{equation*}

1. To convert the weight of a single grain of long grain rice to scientific notation, we must first move the decimal behind the first non-zero digit to obtain \(2.9\text{,}\) which requires that we move the decimal point 2 places. Thus we have:

   \begin{equation*} 0\overbrace{.02}^{2\text{ places}}9\text{ grams} = 2.9 \times 10^{-2}\text{ grams} \end{equation*}

2. The gate pitch of a microprocessor is:

   \begin{equation*} 0\overbrace{.00000001}^{8 \text{ places}}4\text{ meters} = 1.4 \times 10^{-8}\text{ meters} \end{equation*}

1. To convert a download speed of \(7.53\times 10^{-3}\) Gigabyte per second to decimal notation, we will move the decimal point \(3\) places to the left and include the appropriate number of zeros:

   \begin{equation*} 7.53\times 10^{-3}\text{ Gigabyte per second} = 0\overbrace{.007}^{3\text{ places}}53\text{ Gigabyte per second} \end{equation*}

2. The weight of a poppy seed is:

   \begin{equation*} 3 \times 10^{-7}\text{ kilograms} = 0\overbrace{.0000003}^{7 \text{ places}}\text{ kilograms} \end{equation*}

1. The number \(7 \times 10^{1.9}\) is not in scientific notation. The exponent on the 10 is required to be an integer and \(1.9\) is not.

2. The number \(2.6 \times 10^{-31}\) is in scientific notation.

3. The number \(10 \times 7^4\) is not in scientific notation. The base must be \(10\text{,}\) not \(7\text{.}\)

4. The number \(0.93 \times 10^{3}\) is not in scientific notation. The coefficient of the 10 must be between \(1\) (inclusive) and \(10\text{.}\)

5. The number \(4.2 \times 10^0\) is in scientific notation.

6. The number \(12.5 \times 10^{-6}\) is not in scientific notation. The coefficient of the 10 must be between \(1\) (inclusive) and \(10\text{.}\)

We've been asked to answer the same question, but to perform the calculation using two different approaches. In both cases, we'll need to divide the debt by the population.

1. To type this calculation into a calculator, we need to be working a calculator that can handle such large numbers and we have to be extremely careful that we type the correct number of \(0\)s.

   \begin{gather*} \frac{19,600,000,000,000}{323,000,000}\approx 60681.11 \end{gather*}

   The federal debt per capita in the U.S. on September 30th, 2016 was about $60,681.11 per person.

2. To perform this calculation using scientific notation, our work would begin by setting up the quotient \(\dfrac{1.96 \times 10^{13}}{3.23 \times 10^8}\text{.}\) Dividing this quotient follows the same process we did with variable expressions of the same format, such as \(\dfrac{1.96 w^{13}}{3.23 w^8}\text{.}\) In both situations, we'll divide the coefficients and then use exponent rules to simplify the powers.

   \begin{align*} \frac{1.96 \times 10^{13}}{3.23 \times 10^8} \amp= \frac{1.96 }{3.23} \times\frac{10^{13}}{ 10^8} \\ \amp\approx 0.6068111 \times 10^5 \\ \amp\approx 60681.11 \end{align*}

   The federal debt per capita in the U.S. on September 30th, 2016 was about $60,681.11 per person.

Both calculations give us the same answer, but the calculation relying upon scientific notation has less room for error and allows us to perform the calculation as two smaller steps.

Simplifying each of the variable expressions is fundamentally the same process as multiplying or dividing the scientific notation expressions. We will simplify the coefficients as one step and then simplify the exponential factors as a separate step.

1. \begin{align*} \left( 2w^5 \right)\left( 3w^4 \right) \amp= \left( 2\cdot 3 \right)\left( w^5\cdot w^4 \right) \\ \amp= 6 w^{9} \end{align*}

2. \begin{align*} \left( 2\times 10^5 \right)\left( 3\times10^4 \right) \amp= \left( 2\times 3 \right)\times \left(10^5 \times 10^4 \right) \\ \amp= 6 \times 10^{9} \end{align*}

3. \begin{align*} \dfrac{8b^{17}}{4b^2} \amp= \dfrac{8}{4} \cdot \dfrac{b^{17}}{b^2} \\ \amp= 2 \cdot b^{15} \end{align*}

4. \begin{align*} \dfrac{8 \times 10^{17}}{4\times 10^2} \amp= \dfrac{8}{4} \times \dfrac{10^{17}}{10^2} \\ \amp= 2 \times 10^{15} \end{align*}

Again, we'll separate out the work for the coefficients from the work for the powers of ten. If the resulting coefficient is not between one and 10, we'll need to adjust that coefficient to put it into scientific notaiton.

1. \begin{align*} \left( 8 \times 10^{21} \right)\left( 2 \times 10^{-7} \right) \amp= \left( 8 \times 2 \right)\times\left( 10^{21} \times 10^{-7} \right) \\ \amp= \highlight{16} \times 10^{14} \\ \amp= \highlight{1.6\times 10^1} \times 10^{14} \\ \amp= 1.6 \times 10^{15} \end{align*}

   We need to remember to apply the <<Unresolved xref, reference "exponent-rules-product"; check spelling or use "provisional" attribute>> for exponents to the powers of ten.

2. \begin{align*} \dfrac{ 8 \times 10^{21} }{ 2 \times 10^{-7} } \amp= \dfrac{ 8 }{ 2 } \times \dfrac{ 10^{21} }{ 10^{-7} } \\ \amp= 4 \times 10^{28} \end{align*}

   We had to apply the quotient rule for exponents to the powers of ten and to remember that \(21-(-7) = 21+7\text{.}\)

3. \begin{align*} \left( 2 \times 10^{-6} \right)\left( 8 \times 10^{-19} \right) \amp= \left( 2 \times 8 \right)\times \left( 10^{-6} \times 10^{-19} \right) \\ \amp= \highlight{16} \times 10^{-25} \\ \amp= \highlight{1.6\times 10^1} \times 10^{-25} \\ \amp= 1.6 \times 10^{-24} \end{align*}

4. \begin{align*} \dfrac{ 2 \times 10^{-6} }{ 8 \times 10^{-19} } \amp= \dfrac{ 2 }{ 8 }\times\dfrac{ 10^{-6} }{ 10^{-19} } \\ \amp= \highlight{0.25} \times 10^{13} \\ \amp= \highlight{2.5\times 10^{-1}} \times 10^{13} \\ \amp= 2.5 \times 10^{12} \end{align*}