1.2 Exponents and scientific notation (Page 4/9)

College algebra

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

$\frac{{(−3 t)}^{2}}{{(−3 t)}^{8}}$
$\frac{f^{47}}{f^{49} \cdot f}$
$\frac{2 k^{4}}{5 k^{7}}$

$\frac{1}{{(−3 t)}^{6}}$
$\frac{1}{f^{3}}$
$\frac{2}{5 k^{3}}$

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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.

$b^{2} \cdot b^{−8}$
${(- x)}^{5} \cdot {(- x)}^{−5}$
$\frac{−7 z}{{(−7 z)}^{5}}$

$b^{2} \cdot b^{−8} = b^{2 - 8} = b^{−6} = \frac{1}{b^{6}}$
${(- x)}^{5} \cdot {(- x)}^{−5} = {(- x)}^{5 - 5} = {(- x)}^{0} = 1$
$\frac{−7 z}{{(−7 z)}^{5}} = \frac{{(−7 z)}^{1}}{{(−7 z)}^{5}} = {(−7 z)}^{1 - 5} = {(−7 z)}^{−4} = \frac{1}{{(−7 z)}^{4}}$

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Try It

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

$t^{−11} \cdot t^{6}$
$\frac{25^{12}}{25^{13}}$

$t^{−5} = \frac{1}{t^{5}}$
$\frac{1}{25}$

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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 ${(p q)}^{3} .$ We begin by using the associative and commutative properties of multiplication to regroup the factors.

\begin{matrix} {(p q)}^{3} & = & \overset{3 factors}{\overset{}{(p q) \cdot (p q) \cdot (p q)}} \\ = & p \cdot q \cdot p \cdot q \cdot p \cdot q \\ = & \overset{3 factors}{\overset{}{p \cdot p \cdot p}} \cdot \overset{3 factors}{\overset{}{q \cdot q \cdot q}} \\ = & p^{3} \cdot q^{3} \end{matrix}

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

A General Note

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

{(a b)}^{n} = a^{n} b^{n}

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.

${(a b^{2})}^{3}$
${(2 t)}^{15}$
${(−2 w^{3})}^{3}$
$\frac{1}{{(−7 z)}^{4}}$
${(e^{−2} f^{2})}^{7}$

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

${(a b^{2})}^{3} = {(a)}^{3} \cdot {(b^{2})}^{3} = a^{1 \cdot 3} \cdot b^{2 \cdot 3} = a^{3} b^{6}$
$2 t^{15} = {(2)}^{15} \cdot {(t)}^{15} = 2^{15} t^{15} = 32, 768 t^{15}$
${(−2 w^{3})}^{3} = {(−2)}^{3} \cdot {(w^{3})}^{3} = −8 \cdot w^{3 \cdot 3} = −8 w^{9}$
$\frac{1}{{(−7 z)}^{4}} = \frac{1}{{(−7)}^{4} \cdot {(z)}^{4}} = \frac{1}{2, 401 z^{4}}$
${(e^{−2} f^{2})}^{7} = {(e^{−2})}^{7} \cdot {(f^{2})}^{7} = e^{−2 \cdot 7} \cdot f^{2 \cdot 7} = e^{−14} f^{14} = \frac{f^{14}}{e^{14}}$

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Try It

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

${(g^{2} h^{3})}^{5}$
${(5 t)}^{3}$
${(−3 y^{5})}^{3}$
$\frac{1}{{(a^{6} b^{7})}^{3}}$
${(r^{3} s^{−2})}^{4}$

$g^{10} h^{15}$
$125 t^{3}$
$−27 y^{15}$
$\frac{1}{a^{18} b^{21}}$
$\frac{r^{12}}{s^{8}}$

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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^{−2} f^{2})}^{7} = \frac{f^{14}}{e^{14}}

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

\begin{matrix} {(e^{−2} f^{2})}^{7} & = & {(\frac{f^{2}}{e^{2}})}^{7} \\ = & \frac{f^{14}}{e^{14}} \end{matrix}

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

\begin{matrix} {(e^{- 2} f^{2})}^{7} & = & {(\frac{f^{2}}{e^{2}})}^{7} \\ = & \frac{{(f^{2})}^{7}}{{(e^{2})}^{7}} \\ = & \frac{f^{2 \cdot 7}}{e^{2 \cdot 7}} \\ = & \frac{f^{14}}{e^{14}} \end{matrix}

A General Note

The power of a quotient rule of exponents

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

{(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}

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.

${(\frac{4}{z^{11}})}^{3}$
${(\frac{p}{q^{3}})}^{6}$
${(\frac{−1}{t^{2}})}^{27}$
${(j^{3} k^{−2})}^{4}$
${(m^{−2} n^{−2})}^{3}$

${(\frac{4}{z^{11}})}^{3} = \frac{{(4)}^{3}}{{(z^{11})}^{3}} = \frac{64}{z^{11 \cdot 3}} = \frac{64}{z^{33}}$
${(\frac{p}{q^{3}})}^{6} = \frac{{(p)}^{6}}{{(q^{3})}^{6}} = \frac{p^{1 \cdot 6}}{q^{3 \cdot 6}} = \frac{p^{6}}{q^{18}}$
${(\frac{−1}{t^{2}})}^{27} = \frac{{(−1)}^{27}}{{(t^{2})}^{27}} = \frac{−1}{t^{2 \cdot 27}} = \frac{−1}{t^{54}} = - \frac{1}{t^{54}}$
${(j^{3} k^{−2})}^{4} = {(\frac{j^{3}}{k^{2}})}^{4} = \frac{{(j^{3})}^{4}}{{(k^{2})}^{4}} = \frac{j^{3 \cdot 4}}{k^{2 \cdot 4}} = \frac{j^{12}}{k^{8}}$
${(m^{−2} n^{−2})}^{3} = {(\frac{1}{m^{2} n^{2}})}^{3} = \frac{{(1)}^{3}}{{(m^{2} n^{2})}^{3}} = \frac{1}{{(m^{2})}^{3} {(n^{2})}^{3}} = \frac{1}{m^{2 \cdot 3} \cdot n^{2 \cdot 3}} = \frac{1}{m^{6} n^{6}}$

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Questions & Answers

if three forces F1.f2 .f3 act at a point on a Cartesian plane in the daigram .....so if the question says write down the x and y components ..... I really don't understand

Syamthanda Reply

hey , can you please explain oxidation reaction & redox ?

Boitumelo Reply

hey , can you please explain oxidation reaction and redox ?

Boitumelo

for grade 12 or grade 11?

Sibulele

the value of V1 and V2

Tumelo Reply

advantages of electrons in a circuit

Rethabile Reply

we're do you find electromagnetism past papers

Ntombifuthi

what a normal force

Tholulwazi Reply

it is the force or component of the force that the surface exert on an object incontact with it and which acts perpendicular to the surface

Sihle

what is physics?

Petrus Reply

what is the half reaction of Potassium and chlorine

Anna Reply

how to calculate coefficient of static friction

Lisa Reply

how to calculate static friction

Lisa

How to calculate a current

Tumelo

how to calculate the magnitude of horizontal component of the applied force

Mogano

How to calculate force

Monambi

a structure of a thermocouple used to measure inner temperature

Anna Reply

a fixed gas of a mass is held at standard pressure temperature of 15 degrees Celsius .Calculate the temperature of the gas in Celsius if the pressure is changed to 2×10 to the power 4

Amahle Reply

How is energy being used in bonding?

Raymond Reply

what is acceleration

Syamthanda Reply

a rate of change in velocity of an object whith respect to time

Khuthadzo

how can we find the moment of torque of a circular object

Kidist

Acceleration is a rate of change in velocity.

Justice

t =r×f

Khuthadzo

how to calculate tension by substitution

Precious Reply

Shongi

Leago

use fnet method. how many obects are being calculated ?

Khuthadzo

khuthadzo hii

Hulisani

how to calculate acceleration and tension force

Lungile Reply

you use Fnet equals ma , newtoms second law formula

Masego

please help me with vectors in two dimensions

Mulaudzi Reply

how to calculate normal force

Mulaudzi

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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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