1.2 Exponents and scientific notation (Page 5/9)

College algebra

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

${(\frac{b^{5}}{c})}^{3}$
${(\frac{5}{u^{8}})}^{4}$
${(\frac{−1}{w^{3}})}^{35}$
${(p^{−4} q^{3})}^{8}$
${(c^{−5} d^{−3})}^{4}$

$\frac{b^{15}}{c^{3}}$
$\frac{625}{u^{32}}$
$\frac{−1}{w^{105}}$
$\frac{q^{24}}{p^{32}}$
$\frac{1}{c^{20} d^{12}}$

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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.

Simplifying exponential expressions

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

${(6 m^{2} n^{−1})}^{3}$
$17^{5} \cdot 17^{−4} \cdot 17^{−3}$
${(\frac{u^{−1} v}{v^{−1}})}^{2}$
$(−2 a^{3} b^{−1}) (5 a^{−2} b^{2})$
${(x^{2} \sqrt{2})}^{4} {(x^{2} \sqrt{2})}^{−4}$
$\frac{{(3 w^{2})}^{5}}{{(6 w^{−2})}^{2}}$

$\begin{matrix} {(6 m^{2} n^{−1})}^{3} & = & {(6)}^{3} {(m^{2})}^{3} {(n^{−1})}^{3} & The power of a product rule \\ = & 6^{3} m^{2 \cdot 3} n^{−1 \cdot 3} & The power rule \\ = & 216 m^{6} n^{−3} & Simplify . \\ = & \frac{216 m^{6}}{n^{3}} & The negative exponent rule \end{matrix}$
$\begin{matrix} 17^{5} \cdot 17^{−4} \cdot 17^{−3} & = & 17^{5 - 4 - 3} & The product rule \\ = & 17^{−2} & Simplify . \\ = & \frac{1}{17^{2}} or \frac{1}{289} & The negative exponent rule \end{matrix}$
$\begin{matrix} {(\frac{u^{−1} v}{v^{−1}})}^{2} & = & \frac{{(u^{−1} v)}^{2}}{{(v^{−1})}^{2}} & The power of a quotient rule \\ = & \frac{u^{−2} v^{2}}{v^{−2}} & The power of a product rule \\ = & u^{−2} v^{2 - (−2)} & The quotient rule \\ = & u^{−2} v^{4} & Simplify . \\ = & \frac{v^{4}}{u^{2}} & The negative exponent rule \end{matrix}$
$\begin{matrix} (−2 a^{3} b^{- 1}) (5 a^{−2} b^{2}) & = & −2 \cdot 5 \cdot a^{3} \cdot a^{−2} \cdot b^{−1} \cdot b^{2} & Commutative and associative laws of multiplication \\ = & −10 \cdot a^{3 - 2} \cdot b^{−1 + 2} & The product rule \\ = & −10 a b & Simplify . \end{matrix}$
$\begin{matrix} {(x^{2} \sqrt{2})}^{4} {(x^{2} \sqrt{2})}^{−4} & = & {(x^{2} \sqrt{2})}^{4 - 4} & The product rule \\ = & {(x^{2} \sqrt{2})}^{0} & Simplify . \\ = & 1 & The zero exponent rule \end{matrix}$
$\begin{matrix} \frac{{(3 w^{2})}^{5}}{{(6 w^{−2})}^{2}} & = & \frac{{(3)}^{5} \cdot {(w^{2})}^{5}}{{(6)}^{2} \cdot {(w^{−2})}^{2}} & The power of a product rule \\ = & \frac{3^{5} w^{2 \cdot 5}}{6^{2} w^{−2 \cdot 2}} & The power rule \\ = & \frac{243 w^{10}}{36 w^{−4}} & Simplify . \\ = & \frac{27 w^{10 - (−4)}}{4} & The quotient rule and reduce fraction \\ = & \frac{27 w^{14}}{4} & Simplify . \end{matrix}$

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Try It

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

${(2 u v^{- 2})}^{−3}$
$x^{8} \cdot x^{−12} \cdot x$
${(\frac{e^{2} f^{- 3}}{f^{−1}})}^{2}$
$(9 r^{−5} s^{3}) (3 r^{6} s^{−4})$
${(\frac{4}{9} t w^{−2})}^{−3} {(\frac{4}{9} t w^{−2})}^{3}$
$\frac{{(2 h^{2} k)}^{4}}{{(7 h^{−1} k^{2})}^{2}}$

$\frac{v^{6}}{8 u^{3}}$
$\frac{1}{x^{3}}$
$\frac{e^{4}}{f^{4}}$
$\frac{27 r}{s}$
$1$
$\frac{16 h^{10}}{49}$

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Using scientific notation

Recall at the beginning of the section that we found the number $1.3 \times 10^{13}$ when describing bits of information in digital images. Other extreme numbers include the width of a human hair, which is about 0.00005 m, and the radius of an electron, which is about 0.00000000000047 m. How can we effectively work read, compare, and calculate with numbers such as these?

A shorthand method of writing very small and very large numbers is called scientific notation , in which we express numbers in terms of exponents of 10. To write a number in scientific notation, move the decimal point to the right of the first digit in the number. Write the digits as a decimal number between 1 and 10. Count the number of places n that you moved the decimal point. Multiply the decimal number by 10 raised to a power of n . If you moved the decimal left as in a very large number, $n$ is positive. If you moved the decimal right as in a small large number, $n$ is negative.

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

The number 2,780,418 is written with an arrow extending to another number: 2.780418. An arrow tracking the movement of the decimal point runs underneath the number. Above the number a label on the number reads: 6 places left.

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 \times 10^{6}

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.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?

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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance

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Can you compute that for me. Ty

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Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes

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A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity

Krampah Reply

2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.

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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s

Samuel Reply

can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?

Joseph Reply

Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time

Joseph

Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing

Joseph

"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true

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answer

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progressive wave

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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?

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Source: OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3

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