Section 2.3: Properties of Exponents and Logarithms

Section 2.3:  Part 1

This content comes directly from OpenStax’s textbook Algebra and Trigonometry Section 1.2 Exponents and Scientific Notation.

Access this resource for free at https://openstax.org/books/algebra-and-trigonometry-2e/pages/1-2-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.

Let’s Get Started…

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[latex]\,{x}^{3}\cdot {x}^{4}.\,[/latex]Both terms have the same base, x, but they are raised to different exponents. Expand each expression, and then rewrite the resulting expression.

[latex]\begin{array}{ccc}\hfill {x}^{3}\cdot {x}^{4}& =& \stackrel{3\text{ factors}}{\stackrel{}{x\cdot x\cdot x}}\cdot \stackrel{4\text{ factors}}{\stackrel{}{x\cdot x\cdot x\cdot x}}\hfill \\ & =& \stackrel{7\text{ factors}}{\stackrel{}{x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x}}\hfill \\ & =& {x}^{7}\hfill \end{array}[/latex]

 

The result is that[latex]\,{x}^{3}\cdot {x}^{4}={x}^{3+4}={x}^{7}.[/latex]

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.

[latex]{a}^{m}\cdot {a}^{n}={a}^{m+n}[/latex]

Now consider an example with real numbers.

[latex]{2}^{3}\cdot {2}^{4}={2}^{3+4}={2}^{7}[/latex]

We can always check that this is true by simplifying each exponential expression. We find that[latex]\,{2}^{3}\,[/latex]is 8,[latex]\,{2}^{4}\,[/latex]is 16, and[latex]\,{2}^{7}\,[/latex]is 128. The product[latex]\,8\cdot 16\,[/latex]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[latex]\,a\,[/latex]and natural numbers[latex]\,m\,[/latex]and[latex]\,n,[/latex]the product rule of exponents states that

[latex]{a}^{m}\cdot {a}^{n}={a}^{m+n}[/latex].

 

EXAMPLE 1

Using the Product Rule

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

  1. [latex]{t}^{5}\cdot {t}^{3}[/latex]
  2. [latex]{\left(-3\right)}^{5}\cdot \left(-3\right)[/latex]
  3. [latex]{x}^{2}\cdot {x}^{5}\cdot {x}^{3}[/latex]

 

Show/Hide Solution

 

image  Try It #1

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

  1. [latex]{k}^{6}\cdot {k}^{9}[/latex]
  2. [latex]{\left(\frac{2}{y}\right)}^{4}\cdot \left(\frac{2}{y}\right)[/latex]
  3. [latex]{t}^{3}\cdot {t}^{6}\cdot {t}^{5}[/latex]

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[latex]\,\frac{{y}^{m}}{{y}^{n}},[/latex]where[latex]\,m>n.\,[/latex]Consider the example[latex]\,\frac{{y}^{9}}{{y}^{5}}.\,[/latex]Perform the division by canceling common factors.

[latex]\begin{array}{ccc}\hfill \frac{{y}^{9}}{{y}^{5}}& =& \frac{y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y\cdot y}{y\cdot y\cdot y\cdot y\cdot y}\hfill \\ & =& \frac{\cancel{y}\cdot \cancel{y}\cdot \cancel{y}\cdot \cancel{y}\cdot \cancel{y}\cdot y\cdot y\cdot y\cdot y}{\cancel{y}\cdot \cancel{y}\cdot \cancel{y}\cdot \cancel{y}\cdot \cancel{y}}\hfill \\ & =& \frac{y\cdot y\cdot y\cdot y}{1}\hfill \\ & =& {y}^{4}\hfill \end{array}[/latex]

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

[latex]\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}[/latex]

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

[latex]\frac{{y}^{9}}{{y}^{5}}={y}^{9-5}={y}^{4}[/latex]

For the time being, we must be aware of the condition[latex]\,m>n.\,[/latex]Otherwise, the difference[latex]\,m-n\,[/latex]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[latex]\,a\,[/latex]and natural numbers[latex]\,m\,[/latex]and[latex]\,n,[/latex]such that[latex]\,m>n,[/latex]the quotient rule of exponents states that

[latex]\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}[/latex].

 

EXAMPLE 2

Using the Quotient Rule

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

  1. [latex]\frac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}[/latex]
  2. [latex]\frac{{t}^{23}}{{t}^{15}}[/latex]
  3. [latex]\frac{{\left(z\sqrt{2}\right)}^{5}}{z\sqrt{2}}[/latex]

 

Show/Hide Solution

 

image  Try It #2

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

  1. [latex]\frac{{s}^{75}}{{s}^{68}}[/latex]
  2. [latex]\frac{{\left(-3\right)}^{6}}{-3}[/latex]
  3. [latex]\frac{{\left(e{f}^{2}\right)}^{5}}{{\left(e{f}^{2}\right)}^{3}}[/latex]

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[latex]\,{\left({x}^{2}\right)}^{3}.\,[/latex]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.

[latex]\begin{array}{ccc}\hfill {\left({x}^{2}\right)}^{3}& =& \stackrel{3\text{ factors}}{\stackrel{}{\left({x}^{2}\right)\cdot \left({x}^{2}\right)\cdot \left({x}^{2}\right)}}\hfill \\ & =& \hfill \stackrel{3\text{ factors}}{\stackrel{}{\left(\stackrel{2\text{ factors}}{\overbrace{x\cdot x}}\right)\cdot \left(\stackrel{2\text{ factors}}{\overbrace{x\cdot x}}\right)\cdot \left(\stackrel{2\text{ factors}}{\overbrace{x\cdot x}}\right)}}\\ & =& x\cdot x\cdot x\cdot x\cdot x\cdot x\hfill \\ & =& {x}^{6}\hfill \end{array}[/latex]

The exponent of the answer is the product of the exponents:[latex]\,{\left({x}^{2}\right)}^{3}={x}^{2\cdot 3}={x}^{6}.\,[/latex]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.

[latex]{\left({a}^{m}\right)}^{n}={a}^{m\cdot n}[/latex]

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.

[latex]\begin{array}{cccccccccc}& & \text{Product Rule}\hfill & & & & & \text{ Power Rule}\hfill & & \\ \hfill {5}^{3}\cdot {5}^{4}& =& {5}^{3+4}\hfill & =& {5}^{7}\hfill & \phantom{\rule{1em}{0ex}}\text{but}\phantom{\rule{1em}{0ex}}& \hfill \text{ }{\left({5}^{3}\right)}^{4}& =& {5}^{3\cdot 4}\hfill & =& {5}^{12}\hfill \\ \hfill {x}^{5}\cdot {x}^{2}& =& {x}^{5+2}\hfill & =& {x}^{7}\hfill & \phantom{\rule{1em}{0ex}}\text{but}\phantom{\rule{1em}{0ex}}& \hfill {\left({x}^{5}\right)}^{2}& =& {x}^{5\cdot 2}\hfill & =\hfill & {x}^{10}\hfill \\ \hfill {\left(3a\right)}^{7}\cdot {\left(3a\right)}^{10}& =& {\left(3a\right)}^{7+10}\hfill & =& {\left(3a\right)}^{17}\hfill & \phantom{\rule{1em}{0ex}}\text{but}\phantom{\rule{1em}{0ex}}& \hfill {\left({\left(3a\right)}^{7}\right)}^{10}& =& {\left(3a\right)}^{7\cdot 10}\hfill & =& {\left(3a\right)}^{70}\hfill \end{array}[/latex]

 

The Power Rule of Exponents

For any real number[latex]\,a\,[/latex]and positive integers[latex]\,m\,[/latex]and[latex]\,n,[/latex]the power rule of exponents states that

[latex]{\left({a}^{m}\right)}^{n}={a}^{m\cdot n}[/latex].

 

EXAMPLE 3

Using the Power Rule

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

  1. [latex]{\left({x}^{2}\right)}^{7}[/latex]
  2. [latex]{\left({\left(2t\right)}^{5}\right)}^{3}[/latex]
  3. [latex]{\left({\left(-3\right)}^{5}\right)}^{11}[/latex]

 

Show/Hide Solution

 

image  Try It #3

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

  1. [latex]{\left({\left(3y\right)}^{8}\right)}^{3}[/latex]
  2. [latex]{\left({t}^{5}\right)}^{7}[/latex]
  3. [latex]{\left({\left(-g\right)}^{4}\right)}^{4}[/latex]

Using the Zero Exponent Rule of Exponents

Return to the quotient rule. We made the condition that[latex]\,m>n\,[/latex]so that the difference[latex]\,m-n\,[/latex]would never be zero or negative. What would happen if[latex]\,m=n?[/latex]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.

[latex]\frac{{t}^{8}}{{t}^{8}}=\frac{\cancel{{t}^{8}}}{\cancel{{t}^{8}}}=1[/latex]

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

[latex]\frac{{t}^{8}}{{t}^{8}}={t}^{8-8}={t}^{0}[/latex]

If we equate the two answers, the result is[latex]\,{t}^{0}=1.\,[/latex]This is true for any nonzero real number, or any variable representing a real number.

[latex]{a}^{0}=1[/latex]

The sole exception is the expression[latex]\,{0}^{0}.\,[/latex]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[latex]\,a,[/latex]the zero exponent rule of exponents states that

[latex]{a}^{0}=1[/latex].

 

EXAMPLE 4

Using the Zero Exponent Rule

Simplify each expression using the zero exponent rule of exponents.

  1. [latex]\frac{{c}^{3}}{{c}^{3}}[/latex]
  2. [latex]\frac{-3{x}^{5}}{{x}^{5}}[/latex]
  3. [latex]\frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}[/latex]
  4. [latex]\frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}[/latex]

 

Show/Hide Solution

 

image  Try It #4

Simplify each expression using the zero exponent rule of exponents.

  1. [latex]\frac{{t}^{7}}{{t}^{7}}[/latex]
  2. [latex]\frac{{\left(d{e}^{2}\right)}^{11}}{2{\left(d{e}^{2}\right)}^{11}}[/latex]
  3. [latex]\frac{{w}^{4}\cdot {w}^{2}}{{w}^{6}}[/latex]
  4. [latex]\frac{{t}^{3}\cdot {t}^{4}}{{t}^{2}\cdot {t}^{5}}[/latex]

Using the Negative Rule of Exponents

Another useful result occurs if we relax the condition that[latex]\,m>n\,[/latex]in the quotient rule even further. For example, can we simplify[latex]\,\frac{{h}^{3}}{{h}^{5}}?\,[/latex]When[latex]\,m[/latex] < [latex]n[/latex]—that is, where the difference[latex]\,m-n\,[/latex]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,[latex]\,\frac{{h}^{3}}{{h}^{5}}.[/latex]

[latex]\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}[/latex]

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

[latex]\begin{array}{ccc}\hfill \frac{{h}^{3}}{{h}^{5}}& =& {h}^{3-5}\hfill \\ & =& \text{ }{h}^{-2}\hfill \end{array}[/latex].

Putting the answers together, we have[latex]\,{h}^{-2}=\frac{1}{{h}^{2}}.\,[/latex]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.

[latex]\begin{array}{ccc}{a}^{-n}=\frac{1}{{a}^{n}}& \text{and}& {a}^{n}=\frac{1}{{a}^{-n}}\end{array}[/latex]

We have shown that the exponential expression[latex]\,{a}^{n}\,[/latex]is defined when[latex]\,n\,[/latex]is a natural number, 0, or the negative of a natural number. That means that[latex]\,{a}^{n}\,[/latex]is defined for any integer[latex]\,n.\,[/latex]Also, the product and quotient rules and all of the rules we will look at soon hold for any integer[latex]\,n.[/latex]

The Negative Rule of Exponents

For any nonzero real number[latex]\,a\,[/latex]and natural number[latex]\,n,[/latex]the negative rule of exponents states that

[latex]{a}^{-n}=\frac{1}{{a}^{n}}[/latex].

 

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. [latex]\frac{{\theta }^{3}}{{\theta }^{10}}[/latex]
  2. [latex]\frac{{z}^{2}\cdot z}{{z}^{4}}[/latex]
  3. [latex]\frac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}[/latex]

 

Show/Hide Solution

 

image  Try It #5

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

  1. [latex]\frac{{\left(-3t\right)}^{2}}{{\left(-3t\right)}^{8}}[/latex]
  2. [latex]\frac{{f}^{47}}{{f}^{49}\cdot f}[/latex]
  3. [latex]\frac{2{k}^{4}}{5{k}^{7}}[/latex]

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. [latex]{b}^{2}\cdot {b}^{-8}[/latex]
  2. [latex]{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}[/latex]
  3. [latex]\frac{-7z}{{\left(-7z\right)}^{5}}[/latex]

 

Show/Hide Solution

 

image  Try It #6

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

  1. [latex]{t}^{-11}\cdot {t}^{6}[/latex]
  2. [latex]\frac{{25}^{12}}{{25}^{13}}[/latex]

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[latex]\,{\left(pq\right)}^{3}.\,[/latex]We begin by using the associative and commutative properties of multiplication to regroup the factors.

[latex]\begin{array}{ccc}\hfill {\left(pq\right)}^{3}& =& \stackrel{3\text{ factors}}{\stackrel{}{\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}}{\stackrel{}{p\cdot p\cdot p}}\cdot \stackrel{3\text{ factors}}{\stackrel{}{q\cdot q\cdot q}}\hfill \\ & =& {p}^{3}\cdot {q}^{3}\hfill \end{array}[/latex]

In other words,[latex]\,{\left(pq\right)}^{3}={p}^{3}\cdot {q}^{3}.[/latex]

The Power of a Product Rule of Exponents

For any real numbers[latex]\,a\,[/latex]and[latex]\,b\,[/latex]and any integer[latex]\,n,[/latex]the power of a product rule of exponents states that

[latex]{\left(ab\right)}^{n}={a}^{n}{b}^{n}[/latex].

 

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. [latex]{\left(a{b}^{2}\right)}^{3}[/latex]
  2. [latex]{\left(2t\right)}^{15}[/latex]
  3. [latex]{\left(-2{w}^{3}\right)}^{3}[/latex]
  4. [latex]\frac{1}{{\left(-7z\right)}^{4}}[/latex]
  5. [latex]{\left({e}^{-2}{f}^{2}\right)}^{7}[/latex]

 

Show/Hide Solution

 

image  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. [latex]{\left({g}^{2}{h}^{3}\right)}^{5}[/latex]
  2. [latex]{\left(5t\right)}^{3}[/latex]
  3. [latex]{\left(-3{y}^{5}\right)}^{3}[/latex]
  4. [latex]\frac{1}{{\left({a}^{6}{b}^{7}\right)}^{3}}[/latex]
  5. [latex]{\left({r}^{3}{s}^{-2}\right)}^{4}[/latex]

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.

[latex]{\left({e}^{-2}{f}^{2}\right)}^{7}=\frac{{f}^{14}}{{e}^{14}}[/latex]

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

[latex]\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}[/latex]

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

[latex]\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}[/latex]

 

The Power of a Quotient Rule of Exponents

For any real numbers[latex]\,a\,[/latex]and[latex]\,b\,[/latex]and any integer[latex]\,n,[/latex]the power of a quotient rule of exponents states that

[latex]{\left(\frac{a}{b}\right)}^{n}=\frac{{a}^{n}}{{b}^{n}}[/latex].

 

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. [latex]{\left(\frac{4}{{z}^{11}}\right)}^{3}[/latex]
  2. [latex]{\left(\frac{p}{{q}^{3}}\right)}^{6}[/latex]
  3. [latex]{\left(\frac{-1}{{t}^{2}}\right)}^{27}[/latex]
  4. [latex]{\left({j}^{3}{k}^{-2}\right)}^{4}[/latex]
  5. [latex]{\left({m}^{-2}{n}^{-2}\right)}^{3}[/latex]

 

Show/Hide Solution

 

image  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. [latex]{\left(\frac{{b}^{5}}{c}\right)}^{3}[/latex]
  2. [latex]{\left(\frac{5}{{u}^{8}}\right)}^{4}[/latex]
  3. [latex]{\left(\frac{-1}{{w}^{3}}\right)}^{35}[/latex]
  4. [latex]{\left({p}^{-4}{q}^{3}\right)}^{8}[/latex]
  5. [latex]{\left({c}^{-5}{d}^{-3}\right)}^{4}[/latex]

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. [latex]{\left(6{m}^{2}{n}^{-1}\right)}^{3}[/latex]
  2. [latex]{17}^{5}\cdot {17}^{-4}\cdot {17}^{-3}[/latex]
  3. [latex]{\left(\frac{{u}^{-1}v}{{v}^{-1}}\right)}^{2}[/latex]
  4. [latex]\left(-2{a}^{3}{b}^{-1}\right)\left(5{a}^{-2}{b}^{2}\right)[/latex]
  5. [latex]{\left({x}^{2}\sqrt{2}\right)}^{4}{\left({x}^{2}\sqrt{2}\right)}^{-4}[/latex]
  6. [latex]\frac{{\left(3{w}^{2}\right)}^{5}}{{\left(6{w}^{-2}\right)}^{2}}[/latex]

 

Show/Hide Solution

 

image  Try It #9

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

  1. [latex]{\left(2u{v}^{-2}\right)}^{-3}[/latex]
  2. [latex]{x}^{8}\cdot {x}^{-12}\cdot x[/latex]
  3. [latex]{\left(\frac{{e}^{2}{f}^{-3}}{{f}^{-1}}\right)}^{2}[/latex]
  4. [latex]\left(9{r}^{-5}{s}^{3}\right)\left(3{r}^{6}{s}^{-4}\right)[/latex]
  5. [latex]{\left(\frac{4}{9}t{w}^{-2}\right)}^{-3}{\left(\frac{4}{9}t{w}^{-2}\right)}^{3}[/latex]
  6. [latex]\frac{{\left(2{h}^{2}k\right)}^{4}}{{\left(7{h}^{-1}{k}^{2}\right)}^{2}}[/latex]

Section 2.3 Part 1 Exercises

[Answers to odd problem numbers are provided at the end of the problem set.  Just scroll down!]

[Part 2 about Properties of Logarithms follows!]

Verbal

1. Is[latex]\,{2}^{3}\,[/latex]the same as[latex]\,{3}^{2}?\,[/latex]Explain.

 

2. When can you add two exponents?

 

3. When can you subtract two exponents?

 

4. Explain what a negative exponent does.

 

Numeric

For the following exercises, simplify the given expression. Write answers with positive exponents.

5. [latex]\,{9}^{2}\,[/latex]

 

6. [latex]{15}^{-2}[/latex]

 

7. [latex]{3}^{2}\,×\,{3}^{3}[/latex]

 

8. [latex]{4}^{4}÷4[/latex]

 

9. [latex]{\left({2}^{2}\right)}^{-2}[/latex]

 

10. [latex]{\left(5-8\right)}^{0}[/latex]

 

11. [latex]{11}^{3}÷{11}^{4}[/latex]

 

12. [latex]{6}^{5}\,×\,{6}^{-7}[/latex]

 

13. [latex]{\left({8}^{0}\right)}^{2}[/latex]

 

14. [latex]{5}^{-2}÷{5}^{2}[/latex]

 

For the following exercises, write each expression with a single base. Do not simplify further. Write answers with positive exponents.

15. [latex]{4}^{2}\,×\,{4}^{3}÷{4}^{-4}[/latex]

 

16.[latex]\frac{{6}^{12}}{{6}^{9}}[/latex]

 

17. [latex]{\left({12}^{3}\,×\,12\right)}^{10}[/latex]

 

18. [latex]{10}^{6}÷{\left({10}^{10}\right)}^{-2}[/latex]

 

19. [latex]{7}^{-6}\,×\,{7}^{-3}[/latex]

 

20. [latex]{\left({3}^{3}÷{3}^{4}\right)}^{5}[/latex]

 

Algebraic

For the following exercises, simplify the given expression. Write answers with positive exponents.

21, [latex]\frac{{a}^{3}{a}^{2}}{a}[/latex]

 

22. [latex]\frac{m{n}^{2}}{{m}^{-2}}[/latex]

 

23. [latex]{\left({b}^{3}{c}^{4}\right)}^{2}[/latex]

 

24. [latex]{\left(\frac{{x}^{-3}}{{y}^{2}}\right)}^{-5}[/latex]

 

25. [latex]a{b}^{2}÷{d}^{-3}[/latex]

 

26. [latex]{\left({w}^{0}{x}^{5}\right)}^{-1}[/latex]

 

27. [latex]\frac{{m}^{4}}{{n}^{0}}[/latex]

 

28. [latex]{y}^{-4}{\left({y}^{2}\right)}^{2}[/latex]

 

29. [latex]\frac{{p}^{-4}{q}^{2}}{{p}^{2}{q}^{-3}}[/latex]

 

30. [latex]{\left(l\,×\,w\right)}^{2}[/latex]

 

31. [latex]{\left({y}^{7}\right)}^{3}÷{x}^{14}[/latex]

 

32. [latex]{\left(\frac{a}{{2}^{3}}\right)}^{2}[/latex]

 

33. [latex]{5}^{2}m÷{5}^{0}m[/latex]

 

34. [latex]\frac{{\left(16\sqrt{x}\right)}^{2}}{{y}^{-1}}[/latex]

 

35. [latex]\frac{{2}^{3}}{{\left(3a\right)}^{-2}}[/latex]

 

36. [latex]{\left(m{a}^{6}\right)}^{2}\frac{1}{{m}^{3}{a}^{2}}[/latex]

 

37. [latex]{\left({b}^{-3}c\right)}^{3}[/latex]

 

38. [latex]{\left({x}^{2}{y}^{13}÷{y}^{0}\right)}^{2}[/latex]

 

39. [latex]{\left(9{z}^{3}\right)}^{-2}y[/latex]

 

40. [latex]{9{z}^{3}}^{-2}y[/latex]

 

Extensions

For the following exercises, simplify the given expression. Write answers with positive exponents.

41. [latex]{\left(\frac{{3}^{2}}{{a}^{3}}\right)}^{-2}{\left(\frac{{a}^{4}}{{2}^{2}}\right)}^{2}[/latex]

 

42. [latex]{\left({6}^{2}-24\right)}^{2}÷{\left(\frac{x}{y}\right)}^{-5}[/latex]

 

43. [latex]\frac{{m}^{2}{n}^{3}}{{a}^{2}{c}^{-3}}\cdot \frac{{a}^{-7}{n}^{-2}}{{m}^{2}{c}^{4}}[/latex]

 

44. [latex]{\left(\frac{{x}^{6}{y}^{3}}{{x}^{3}{y}^{-3}}\cdot \frac{{y}^{-7}}{{x}^{-3}}\right)}^{10}[/latex]

 

45. [latex]{\left(\frac{{\left(a{b}^{2}c\right)}^{-3}}{{b}^{-3}}\right)}^{2}[/latex]

 

Answers to Section 2.3 Part 1 Odd Problems

1. No, the two expressions are not the same. An exponent tells how many times you multiply the base. So[latex]\,{2}^{3}\,[/latex]is the same as[latex]\,2×2×2,[/latex]which is 8.[latex]\,{3}^{2}\,[/latex]is the same as[latex]\,3×3,[/latex]which is 9.

 

3. When you have [latex]\frac{{a}^{m}}{{a}^{n}}[/latex].

 

5. 81

 

7. 243

 

9. [latex]\frac{1}{16}[/latex]

 

11. [latex]\frac{1}{11}[/latex]

 

13. 1

 

15. [latex]{4}^{9}[/latex]

 

17. [latex]{12}^{40}[/latex]

 

19. [latex]\frac{1}{{7}^{9}}[/latex]

 

21. [latex]{a}^{4}[/latex]

 

23. [latex]{b}^{6}{c}^{8}[/latex]

 

25. [latex]a{b}^{2}{d}^{3}[/latex]

 

27. [latex]{m}^{4}[/latex]

 

29. [latex]\frac{{q}^{5}}{{p}^{6}}[/latex]

 

31. [latex]\frac{{y}^{21}}{{x}^{14}}[/latex]

 

33. [latex]25[/latex]

 

35. [latex]72{a}^{2}[/latex]

 

37. [latex]\frac{{c}^{3}}{{b}^{9}}[/latex]

 

39. [latex]\frac{y}{81{z}^{6}}[/latex]

 

41. [latex]\frac{{a}^{14}}{1296}[/latex]

 

43. [latex]\frac{n}{{a}^{9}c}[/latex]

 

45. [latex]\frac{1}{{a}^{6}{b}^{6}{c}^{6}}[/latex]

 

Section 2.3:  Part 2

This content comes directly from OpenStax’s textbook Algebra and Trigonometry Section 6.5 Logarithmic Properties

Access this resource for free at https://openstax.org/books/algebra-and-trigonometry-2e/pages/6-5-logarithmic-properties

Learning Objectives

In this section, you will:

  • Use the product rule for logarithms.
  • Use the quotient rule for logarithms.
  • Use the power rule for logarithms.
  • Expand logarithmic expressions.
  • Condense logarithmic expressions.

Let’s Get Started…

Testing of the pH of hydrochloric acid.

Figure 1 The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan)

 

In chemistry, pH is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from 0 to 14. Substances with a pH less than 7 are considered acidic, and substances with a pH greater than 7 are said to be alkaline. Our bodies, for instance, must maintain a pH close to 7.35 in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances:

  • Battery acid: 0.8
  • Stomach acid: 2.7
  • Orange juice: 3.3
  • Pure water: 7 (at 25° C)
  • Human blood: 7.35
  • Fresh coconut: 7.8
  • Sodium hydroxide (lye): 14

To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where[latex]\,a\,[/latex]is the concentration of hydrogen ion in the solution

[latex]\begin{array}{l}\text{pH}=-\mathrm{log}\left(\left[{H}^{+}\right]\right)\hfill \\ \text{ }=\mathrm{log}\left(\frac{1}{\left[{H}^{+}\right]}\right).\hfill \end{array}[/latex]

The equivalence of[latex]\,-\mathrm{log}\left(\left[{H}^{+}\right]\right)\,[/latex]and[latex]\,\mathrm{log}\left(\frac{1}{\left[{H}^{+}\right]}\right)\,[/latex]is one of the logarithm properties we will examine in this section.

Using the Product Rule for Logarithms

Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.

[latex]\begin{array}{l}{\mathrm{log}}_{b}1=0\\ {\mathrm{log}}_{b}b=1\end{array}[/latex]

For example,[latex]\,{\mathrm{log}}_{5}1=0\,[/latex]since[latex]\,{5}^{0}=1.\,[/latex]And[latex]\,{\mathrm{log}}_{5}5=1\,[/latex]since[latex]\,{5}^{1}=5.[/latex]

Next, we have the inverse property.

[latex]\begin{array}{l}\hfill \\ {\mathrm{log}}_{b}\left({b}^{x}\right)=x\hfill \\ \text{ }{b}^{{\mathrm{log}}_{b}x}=x,x>0\hfill \end{array}[/latex]

For example, to evaluate[latex]\,\mathrm{log}\left(100\right),[/latex]we can rewrite the logarithm as[latex]\,{\mathrm{log}}_{10}\left({10}^{2}\right),[/latex] and then apply the inverse property[latex]\,{\mathrm{log}}_{b}\left({b}^{x}\right)=x\,[/latex]to get[latex]\,{\mathrm{log}}_{10}\left({10}^{2}\right)=2.[/latex]

To evaluate[latex]\,{e}^{\mathrm{ln}\left(7\right)},[/latex]we can rewrite the logarithm as[latex]\,{e}^{{\mathrm{log}}_{e}7},[/latex] and then apply the inverse property[latex]\,{b}^{{\mathrm{log}}_{b}x}=x\,[/latex]to get[latex]\,{e}^{{\mathrm{log}}_{e}7}=7.[/latex]

Finally, we have the one-to-one property.

[latex]{\mathrm{log}}_{b}M={\mathrm{log}}_{b}N\,\text{ if and only if}\,\text{ }M=N[/latex]

We can use the one-to-one property to solve the equation[latex]\,{\mathrm{log}}_{3}\left(3x\right)={\mathrm{log}}_{3}\left(2x+5\right)\,[/latex]for[latex]\,x.\,[/latex]Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for[latex]\,x:[/latex]

[latex]\begin{array}{ll}3x=2x+5\,\,\,\,\,\hfill & \text{Set the arguments equal}\text{.}\hfill \\ \,\,\,\,x=5\hfill & \text{Subtract 2}x\text{.}\hfill \end{array}[/latex]

But what about the equation[latex]\,{\mathrm{log}}_{3}\left(3x\right)+{\mathrm{log}}_{3}\left(2x+5\right)=2?\,[/latex]The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation.

Recall that we use the product rule of exponents to combine the product of exponents by adding:[latex]\,{x}^{a}{x}^{b}={x}^{a+b}.\,[/latex]We have a similar property for logarithms, called the product rule for logarithms, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.

Given any real number[latex]\,x\,[/latex]and positive real numbers[latex]\text{ }M,N,[/latex] and[latex]\,b,[/latex] where[latex]\,b\ne 1,[/latex] we will show

[latex]\,{\mathrm{log}}_{b}\left(MN\right)\text{=}{\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right).[/latex]

Let[latex]\,m={\mathrm{log}}_{b}M\,[/latex]and[latex]\,n={\mathrm{log}}_{b}N.\,[/latex]In exponential form, these equations are[latex]\,{b}^{m}=M\,[/latex]and[latex]\,{b}^{n}=N.\,[/latex]It follows that

[latex]\begin{array}{lll}{\mathrm{log}}_{b}\left(MN\right)\hfill & ={\mathrm{log}}_{b}\left({b}^{m}{b}^{n}\right)\,\,\,\,\,\,\hfill & \text{Substitute for }M\text{ and }N.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left({b}^{m+n}\right)\hfill & \text{Apply the product rule for exponents}.\hfill \\ \hfill & =m+n\hfill & \text{Apply the inverse property of logs}.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)\hfill & \text{Substitute for }m\text{ and }n.\hfill \end{array}[/latex]

Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider[latex]\,{\mathrm{log}}_{b}\left(wxyz\right).\,[/latex]Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:

[latex]{\mathrm{log}}_{b}\left(wxyz\right)={\mathrm{log}}_{b}w+{\mathrm{log}}_{b}x+{\mathrm{log}}_{b}y+{\mathrm{log}}_{b}z[/latex].

 

The Product Rule for Logarithms

The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms

[latex]{\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)\text{ for }b>0[/latex]

 

How To

Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms.

  1. Factor the argument completely, expressing each whole number factor as a product of primes.
  2. Write the equivalent expression by summing the logarithms of each factor.

 

EXAMPLE 1

Using the Product Rule for Logarithms

Expand [latex]\,{\mathrm{log}}_{3}\left(30x\left(3x+4\right)\right).[/latex]

 

Show/Hide Solution

 

image  Try It #1

Expand[latex]\,{\mathrm{log}}_{b}\left(8k\right).[/latex]

Using the Quotient Rule for Logarithms

For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to combine the quotient of exponents by subtracting:[latex]\,\frac{{x}^{a}}{{x}^{b}}={x}^{a-b}.\,[/latex]The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.

Given any real number[latex]\,x\,[/latex]and positive real numbers[latex]\,M,[/latex][latex]N,[/latex] and [latex]\,b,[/latex] where[latex]\,b\ne 1,[/latex] we will show

[latex]{\mathrm{log}}_{b}\left(\frac{M}{N}\right)\text{=}{\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right).[/latex]

Let[latex]\,m={\mathrm{log}}_{b}M\,[/latex]and[latex]\,n={\mathrm{log}}_{b}N.\,[/latex]In exponential form, these equations are[latex]\,{b}^{m}=M\,[/latex]and[latex]\,{b}^{n}=N.\,[/latex]It follows that

 

[latex]\begin{array}{lll}{\mathrm{log}}_{b}\left(\frac{M}{N}\right)\hfill & ={\mathrm{log}}_{b}\left(\frac{{b}^{m}}{{b}^{n}}\right)\,\,\,\,\,\hfill & \text{Substitute for }M\text{ and }N.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left({b}^{m-n}\right)\hfill & \text{Apply the quotient rule for exponents}.\hfill \\ \hfill & =m-n\hfill & \text{Apply the inverse property of logs}.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)\hfill & \text{Substitute for }m\text{ and }n.\hfill \end{array}[/latex]

 

For example, to expand[latex]\,\mathrm{log}\left(\frac{2{x}^{2}+6x}{3x+9}\right),[/latex]we must first express the quotient in lowest terms.

Factoring and canceling we get,

[latex]\begin{array}{ll}\mathrm{log}\left(\frac{2{x}^{2}+6x}{3x+9}\right)=\mathrm{log}\left(\frac{2x\left(x+3\right)}{3\left(x+3\right)}\right)\hfill & \text{Factor the numerator and denominator}.\hfill \\ \text{ }=\mathrm{log}\left(\frac{2x}{3}\right)\hfill & \text{Cancel the common factors}.\hfill \end{array}[/latex]

Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.

[latex]\begin{array}{l}\mathrm{log}\left(\frac{2x}{3}\right)=\mathrm{log}\left(2x\right)-\mathrm{log}\left(3\right)\hfill \\ \text{ }=\mathrm{log}\left(2\right)+\mathrm{log}\left(x\right)-\mathrm{log}\left(3\right)\hfill \end{array}[/latex].

 

The Quotient Rule for Logarithms

The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.

[latex]{\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}M-{\mathrm{log}}_{b}N[/latex]

 

How To

Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.

  1. Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.
  2. Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.
  3. Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.

 

EXAMPLE 2

Using the Product Rule for Logarithms

Expand[latex]\,{\mathrm{log}}_{2}\left(\frac{15x\left(x-1\right)}{\left(3x+4\right)\left(2-x\right)}\right).[/latex]

 

Show/Hide Solution

 

image Analysis

There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for[latex]\,x=-\frac{4}{3}\,[/latex]and[latex]\,x=2.\,[/latex]Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that[latex]\,x>0,[/latex][latex]x>1,[/latex][latex]x>-\frac{4}{3},[/latex] and[latex]\,x<2.\,[/latex]Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.

 

image  Try It #2

Expand[latex]\,{\mathrm{log}}_{3}\left(\frac{7{x}^{2}+21x}{7x\left(x-1\right)\left(x-2\right)}\right).[/latex]

Using the Power Rule for Logarithms

We’ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as[latex]\,{x}^{2}?\,[/latex]One method is as follows:

[latex]\begin{array}{ll}{\mathrm{log}}_{b}\left({x}^{2}\right)\hfill & ={\mathrm{log}}_{b}\left(x\cdot x\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}x+{\mathrm{log}}_{b}x\hfill \\ \hfill & =2{\mathrm{log}}_{b}x\hfill \end{array}[/latex]

Notice that we used the product rule for logarithms to find a solution for the example above. By doing so, we have derived the power rule for logarithms, which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,

[latex]\begin{array}{lll}100={10}^{2}\,\,\,\,\,\hfill & \sqrt{3}={3}^{\frac{1}{2}}\,\,\,\,\,\hfill & \frac{1}{e}={e}^{-1}\hfill \end{array}[/latex]

 

The Power Rule for Logarithms

The power rule for logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.

[latex]{\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M[/latex]

 

How To

Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm.

  1. Express the argument as a power, if needed.
  2. Write the equivalent expression by multiplying the exponent times the logarithm of the base.

 

EXAMPLE 3

Expanding a Logarithm with Powers

Expand[latex]\,{\mathrm{log}}_{2}{x}^{5}.[/latex]

 

Show/Hide Solution

 

image  Try It #3

Expand[latex]\,\mathrm{ln}{x}^{2}.\,[/latex]

EXAMPLE 4

Expanding a Logarithm with Powers

Expand[latex]\,{\mathrm{log}}_{3}\left(25\right)\,[/latex]using the power rule for logs.

 

Show/Hide Solution

 

image  Try It #4

Expand[latex]\,\mathrm{ln}\left(\frac{1}{{x}^{2}}\right).[/latex]

EXAMPLE 5

Using the Power Rule in Reverse

Rewrite[latex]\,4\mathrm{ln}\left(x\right)\,[/latex]using the power rule for logs to a single logarithm with a leading coefficient of 1.

 

Show/Hide Solution

 

image  Try It #5

Rewrite[latex]\,2{\mathrm{log}}_{3}4\,[/latex]using the power rule for logs to a single logarithm with a leading coefficient of 1.

Expanding Logarithmic Expressions

Taken together, the product rule, quotient rule, and power rule are often called “laws of logs.” Sometimes we apply more than one rule in order to simplify an expression. For example:

[latex]\begin{array}{ll}{\mathrm{log}}_{b}\left(\frac{6x}{y}\right)\hfill & ={\mathrm{log}}_{b}\left(6x\right)-{\mathrm{log}}_{b}y\hfill \\ \hfill & ={\mathrm{log}}_{b}6+{\mathrm{log}}_{b}x-{\mathrm{log}}_{b}y\hfill \end{array}[/latex]

We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal is a negative power:

[latex]\begin{array}{ll}{\mathrm{log}}_{b}\left(\frac{A}{C}\right)\hfill & ={\mathrm{log}}_{b}\left(A{C}^{-1}\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(A\right)+{\mathrm{log}}_{b}\left({C}^{-1}\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}A+\left(-1\right){\mathrm{log}}_{b}C\hfill \\ \hfill & ={\mathrm{log}}_{b}A-{\mathrm{log}}_{b}C\hfill \end{array}[/latex]

We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.

With practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots—never with addition or subtraction inside the argument of the logarithm.

 

EXAMPLE 6

Expanding Logarithms Using Product, Quotient, and Power Rules

Rewrite[latex]\,\mathrm{ln}\left(\frac{{x}^{4}y}{7}\right)\,[/latex]as a sum or difference of logs.

 

Show/Hide Solution

 

image  Try It #6

Expand[latex]\,\mathrm{log}\left(\frac{{x}^{2}{y}^{3}}{{z}^{4}}\right).[/latex]

EXAMPLE 7

Expanding Logarithms Using Product, Quotient, and Power Rules

Expand[latex]\,\mathrm{log}\left(\sqrt{x}\right).[/latex]

 

Show/Hide Solution

 

image  Try It #7

Expand[latex]\,\mathrm{ln}\left(\sqrt[3]{{x}^{2}}\right).[/latex]

image  Q&A

Can we expand[latex]\,\mathrm{ln}\left({x}^{2}+{y}^{2}\right)?[/latex]

No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.

EXAMPLE 8

Expanding Complex Logarithmic Expressions

Expand[latex]\,{\mathrm{log}}_{6}\left(\frac{64{x}^{3}\left(4x+1\right)}{\left(2x-1\right)}\right).[/latex]

 

Show/Hide Solution

 

image  Try It #8

Expand[latex]\,\mathrm{ln}\left(\frac{\sqrt{\left(x-1\right){\left(2x+1\right)}^{2}}}{\left({x}^{2}-9\right)}\right).[/latex]

Condensing Logarithmic Expressions

We can use the rules of logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined. We will learn later how to change the base of any logarithm before condensing.

 

How To

Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm.

  1. Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.
  2. Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.
  3. Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.

 

EXAMPLE 9

Using the Product and Quotient Rules to Combine Logarithms

Write[latex]\,{\mathrm{log}}_{3}\left(5\right)+{\mathrm{log}}_{3}\left(8\right)-{\mathrm{log}}_{3}\left(2\right)\,[/latex]as a single logarithm.

 

Show/Hide Solution

 

image  Try It #9

Condense[latex]\,\mathrm{log}3-\mathrm{log}4+\mathrm{log}5-\mathrm{log}6.[/latex]

EXAMPLE 10

Condensing Complex Logarithmic Expressions

Condense[latex]\,{\mathrm{log}}_{2}\left({x}^{2}\right)+\frac{1}{2}{\mathrm{log}}_{2}\left(x-1\right)-3{\mathrm{log}}_{2}\left({\left(x+3\right)}^{2}\right).[/latex]

 

Show/Hide Solution

 

EXAMPLE 11

Rewriting as a Single Logarithm

Rewrite[latex]\,2\mathrm{log}x-4\mathrm{log}\left(x+5\right)+\frac{1}{x}\mathrm{log}\left(3x+5\right)\,[/latex]as a single logarithm.

 

Show/Hide Solution

 

image  Try It #10

Rewrite[latex]\,\mathrm{log}\left(5\right)+0.5\mathrm{log}\left(x\right)-\mathrm{log}\left(7x-1\right)+3\mathrm{log}\left(x-1\right)\,[/latex]as a single logarithm.

image  Try It #11

Condense[latex]\,4\left(3\mathrm{log}\left(x\right)+\mathrm{log}\left(x+5\right)-\mathrm{log}\left(2x+3\right)\right).[/latex]

EXAMPLE 12

Applying of the Laws of Logs

Recall that, in chemistry,[latex]\,\text{pH}=-\mathrm{log}\left[{H}^{+}\right].\,[/latex]If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?

 

Show/Hide Solution

 

image  Try It #12

How does the pH change when the concentration of positive hydrogen ions is decreased by half?

Section 2.3 Part 2 Exercises

[Answers to odd problem numbers are provided at the end of the problem set.  Just scroll down!]

Verbal

1. How does the power rule for logarithms help when solving logarithms with the form[latex]\,{\mathrm{log}}_{b}\left(\sqrt[n]{x}\right)?[/latex]

 

2. What does the change-of-base formula do? Why is it useful when using a calculator?

 

Algebraic

For the following exercises, expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

3.[latex]{\mathrm{log}}_{b}\left(7x\cdot 2y\right)[/latex]

 

4. [latex]\mathrm{ln}\left(3ab\cdot 5c\right)[/latex]

 

5. [latex]{\mathrm{log}}_{b}\left(\frac{13}{17}\right)[/latex]

 

6. [latex]{\mathrm{log}}_{4}\left(\frac{\text{ }\frac{x}{z}\text{ }}{w}\right)[/latex]

 

7. [latex]\mathrm{ln}\left(\frac{1}{{4}^{k}}\right)[/latex]

 

8. [latex]{\mathrm{log}}_{2}\left({y}^{x}\right)[/latex]

 

For the following exercises, condense to a single logarithm if possible

9. [latex]\mathrm{ln}\left(7\right)+\mathrm{ln}\left(x\right)+\mathrm{ln}\left(y\right)[/latex]

 

10. [latex]{\mathrm{log}}_{3}\left(2\right)+{\mathrm{log}}_{3}\left(a\right)+{\mathrm{log}}_{3}\left(11\right)+{\mathrm{log}}_{3}\left(b\right)[/latex]

 

11. [latex]{\mathrm{log}}_{b}\left(28\right)-{\mathrm{log}}_{b}\left(7\right)[/latex]

 

12. [latex]\mathrm{ln}\left(a\right)-\mathrm{ln}\left(d\right)-\mathrm{ln}\left(c\right)[/latex]

 

13. [latex]-{\mathrm{log}}_{b}\left(\frac{1}{7}\right)[/latex]

 

14. [latex]\frac{1}{3}\mathrm{ln}\left(8\right)[/latex]

 

For the following exercises, use the properties of logarithms to expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

15. [latex]\mathrm{log}\left(\frac{{x}^{15}{y}^{13}}{{z}^{19}}\right)[/latex]

 

16. [latex]\mathrm{ln}\left(\frac{{a}^{-2}}{{b}^{-4}{c}^{5}}\right)[/latex]

 

17. [latex]\mathrm{log}\left(\sqrt{{x}^{3}{y}^{-4}}\right)[/latex]

 

18. [latex]\mathrm{ln}\left(y\sqrt{\frac{y}{1-y}}\right)[/latex]

 

19. [latex]\mathrm{log}\left({x}^{2}{y}^{3}\sqrt[3]{{x}^{2}{y}^{5}}\right)[/latex]

 

For the following exercises, condense each expression to a single logarithm using the properties of logarithms.

20. [latex]\mathrm{log}\left(2{x}^{4}\right)+\mathrm{log}\left(3{x}^{5}\right)[/latex]

 

21. [latex]\mathrm{ln}\left(6{x}^{9}\right)-\mathrm{ln}\left(3{x}^{2}\right)[/latex]

 

22. [latex]2\mathrm{log}\left(x\right)+3\mathrm{log}\left(x+1\right)[/latex]

 

23. [latex]\mathrm{log}\left(x\right)-\frac{1}{2}\mathrm{log}\left(y\right)+3\mathrm{log}\left(z\right)[/latex]

 

24. [latex]4{\mathrm{log}}_{7}\left(c\right)+\frac{{\mathrm{log}}_{7}\left(a\right)}{3}+\frac{{\mathrm{log}}_{7}\left(b\right)}{3}[/latex]

 

Numeric

For the following exercises, use properties of logarithms to evaluate without using a calculator.

25. [latex]6{\mathrm{log}}_{8}\left(2\right)+\frac{{\mathrm{log}}_{8}\left(64\right)}{3{\mathrm{log}}_{8}\left(4\right)}[/latex]

 

26. [latex]{\mathrm{log}}_{3}\left(\frac{1}{9}\right)-3{\mathrm{log}}_{3}\left(3\right)[/latex]

 

27. [latex]2{\mathrm{log}}_{9}\left(3\right)-4{\mathrm{log}}_{9}\left(3\right)+{\mathrm{log}}_{9}\left(\frac{1}{729}\right)[/latex]

 

Extensions

28. Use the product rule for logarithms to find all[latex]\,x\,[/latex]values such that[latex]\,{\mathrm{log}}_{12}\left(2x+6\right)+{\mathrm{log}}_{12}\left(x+2\right)=2.\,[/latex]Show the steps for solving.

 

29. Use the quotient rule for logarithms to find all[latex]\,x\,[/latex]values such that[latex]\,{\mathrm{log}}_{6}\left(x+2\right)-{\mathrm{log}}_{6}\left(x-3\right)=1.\,[/latex]Show the steps for solving.

 

30. Can the power property of logarithms be derived from the power property of exponents using the equation[latex]\,{b}^{x}=m?\,[/latex]If not, explain why. If so, show the derivation.

 

31. Prove that[latex]\,{\mathrm{log}}_{b}\left(n\right)=\frac{1}{{\mathrm{log}}_{n}\left(b\right)}\,[/latex]for any positive integers[latex]\,b>1\,[/latex]and[latex]\,n>1.[/latex]

 

32. Does[latex]\,{\mathrm{log}}_{81}\left(2401\right)={\mathrm{log}}_{3}\left(7\right)?\,[/latex]Verify the claim algebraically.

 

Answers to Section 2.3 Part 2 Odd Problems

1. Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus,[latex]\,{\mathrm{log}}_{b}\left({x}^{\frac{1}{n}}\right)=\frac{1}{n}{\mathrm{log}}_{b}\left(x\right).[/latex]

 

3. [latex]{\mathrm{log}}_{b}\left(2\right)+{\mathrm{log}}_{b}\left(7\right)+{\mathrm{log}}_{b}\left(x\right)+{\mathrm{log}}_{b}\left(y\right)[/latex]

 

5. [latex]{\mathrm{log}}_{b}\left(13\right)-{\mathrm{log}}_{b}\left(17\right)[/latex]

 

7. [latex]-k\mathrm{ln}\left(4\right)[/latex]

 

9.[latex]\mathrm{ln}\left(7xy\right)[/latex]

 

11. [latex]{\mathrm{log}}_{b}\left(4\right)[/latex]

 

13. [latex]{\text{log}}_{b}\left(7\right)[/latex]

 

15. [latex]15\mathrm{log}\left(x\right)+13\mathrm{log}\left(y\right)-19\mathrm{log}\left(z\right)[/latex]

 

17. [latex]\frac{3}{2}\mathrm{log}\left(x\right)-2\mathrm{log}\left(y\right)[/latex]

 

19. [latex]\frac{8}{3}\mathrm{log}\left(x\right)+\frac{14}{3}\mathrm{log}\left(y\right)[/latex]

 

21. [latex]\mathrm{ln}\left(2{x}^{7}\right)[/latex]

 

23. [latex]\mathrm{log}\left(\frac{x{z}^{3}}{\sqrt{y}}\right)[/latex]

 

25. [latex]3[/latex]

 

27. [latex]-4[/latex]

 

29.  [latex]x=4;\,[/latex]By the quotient rule:[latex]{\mathrm{log}}_{6}\left(x+2\right)-{\mathrm{log}}_{6}\left(x-3\right)={\mathrm{log}}_{6}\left(\frac{x+2}{x-3}\right)=1.[/latex]

Rewriting as an exponential equation and solving for[latex]\,x:[/latex]

[latex]\begin{array}{ll}{6}^{1}\hfill & =\frac{x+2}{x-3}\hfill \\ \,0\hfill & =\frac{x+2}{x-3}-6\hfill \\ \,0\hfill & =\frac{x+2}{x-3}-\frac{6\left(x-3\right)}{\left(x-3\right)}\hfill \\ \,0\hfill & =\frac{x+2-6x+18}{x-3}\hfill \\ \,0\hfill & =\frac{x-4}{x-3}\hfill \\ \text{​}\,x\hfill & =4\hfill \end{array}[/latex]

Checking, we find that[latex]\,{\mathrm{log}}_{6}\left(4+2\right)-{\mathrm{log}}_{6}\left(4-3\right)={\mathrm{log}}_{6}\left(6\right)-{\mathrm{log}}_{6}\left(1\right)\,[/latex]is defined, so[latex]\,x=4.[/latex]

 

31. Let[latex]\,b\,[/latex]and[latex]\,n\,[/latex]be positive integers greater than[latex]\,1.\,[/latex]Then, by the change-of-base formula,

[latex]\,{\mathrm{log}}_{b}\left(n\right)=\frac{{\mathrm{log}}_{n}\left(n\right)}{{\mathrm{log}}_{n}\left(b\right)}=\frac{1}{{\mathrm{log}}_{n}\left(b\right)}.[/latex]

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

Baylor University's Co-requisite Supplement for Calculus I Copyright © 2023 by Amy Graham Goodman is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Share This Book