TBIL-Precal Properties of Logarithms (EL5)

Notice this fact will eventually help us solve logarithmic equations. If we have an equation where each side is expressed as a single logarithm with matching bases (such as

\log_{b} M = \log_{b} N

), then it follows that the arguments (

M

and

N

) are also equal to each other.

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Remark 5.5.5.

Recall that exponential functions and logarithmic functions are inverses. We know that

\log_{b} (b^{k}) = k

and according to the law of exponents, we know that:

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x^{a} \cdot x^{b} = x^{a + b}

\frac{x^{a}}{x^{b}} = x^{a - b}

{(x^{a})}^{b} = x^{a \cdot b}

Consider all these as you move through the activities in this section.

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Activity 5.5.6.

Let’s begin with the law of exponents to see if we can understand the product property of logs. According to the law of exponents, we know that

10^{x} \cdot 10^{y} = 10^{x + y} .

Start with this equation as you move through this activity.

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(a)

Let

a = 10^{x}

and

b = 10^{y} .

How could you rewrite the left side of the equation

10^{x} \cdot 10^{y} ?

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$a + b$
$a - b$
$10^{x + y}$
$a \cdot b$

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Answer.

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(b)

Recall from Fact 5.5.3 that

\log_{b} M = \log_{b} N

if and only if

M = N .

Use this property to apply the logarithm to both sides of the rewritten equation from part (a). What is that equation?

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$\log_{10} (a + b) = \log_{10} (10^{x + y})$
$\log_{10} (a \cdot b) = \log_{10} (10^{x + y})$
$\log_{10} (a \cdot b) = \log_{10} (10^{a + b})$
$\log_{10} (a + b) = \log_{10} (10^{a + b})$

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Answer.

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(c)

Knowing that

\log_{b} (b^{k}) = k,

how could you simplify the right side of the equation?

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$\log_{10} (a + b)$
$\log_{10} (x + y)$
$a + b$
$x + y$

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Answer.

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(d)

Recall in part (a), we defined

10^{x} = a

and

10^{y} = b .

What would these look like in logarithmic form?

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$\log_{10} a = x$
$\log_{x} a = 10$
$\log_{10} b = y$
$\log_{y} b = 10$

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Answer.

A and C

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(e)

Using your solutions in part (d), how can we rewrite the right side of the equation?

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$10^{a + b}$
$\log_{10} a - \log_{10} b$
$\log_{10} a + \log_{10} b$
$10^{x} + 10^{y}$

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Answer.

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(f)

Combining parts (a) and (d), which equation represents

10^{x} \cdot 10^{y} = 10^{x + y}

in terms of logarithms?

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$\log_{10} (a + b) = 10^{a + b}$
$\log_{10} (a \cdot b) = \log_{10} a - \log_{10} b$
$\log_{10} (a \cdot b) = \log_{10} a + \log_{10} b$
$\log_{10} (a \cdot b) = 10^{x} + 10^{y}$

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Answer.

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Activity 5.5.7.

According to the law of exponents, we know that

\frac{10^{x}}{10^{y}} = 10^{x - y} .

Start with this equation as you move through this activity.

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(a)

Let

a = 10^{x}

and

b = 10^{y} .

How could you rewrite the left side of the equation

\frac{10^{x}}{10^{y}} ?

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$a + b$
$a - b$
$10^{x + y}$
$a \cdot b$

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Answer.

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(b)

Use the one-to-one property of logarithms to apply the logarithm to both sides of the rewritten equation from part (a). What is that equation?

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$\log_{10} (a - b) = \log_{10} (10^{x - y})$
$\log_{10} (\frac{a}{b}) = \log_{10} (10^{x - y})$
$\log_{10} (\frac{a}{b}) = \log_{10} (10^{a + b})$
$\log_{10} (a - b) = \log_{10} (10^{a - b})$

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Answer.

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(c)

Knowing that

\log_{b} (b^{k}) = k,

how could you simplify the right side of the equation?

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$\log_{10} (a - b)$
$\log_{10} (x - y)$
$x - y$
$a - b$

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Answer.

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(d)

Recall in part (a), we defined

10^{x} = a

and

10^{y} = b .

What would these look like in logarithmic form?

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$\log_{10} a = x$
$\log_{x} a = 10$
$\log_{10} b = y$
$\log_{y} b = 10$

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Answer.

A and C

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(e)

Using your solutions in part (d), how can we rewrite the right side of the equation?

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$10^{a + b}$
$\log_{10} a - \log_{10} b$
$\log_{10} a - \log_{10} b$
$10^{x - y}$

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Answer.

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(f)

Combining parts (a) and (d), which equation represents

\frac{10^{x}}{10^{y}} = 10^{x - y}

in terms of logarithms?

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$\log_{10} (a - b) = 10^{a + b}$
$\log_{10} (a - b) = \log_{10} a - \log_{10} b$
$\log_{10} (\frac{a}{b}) = \log_{10} a - \log_{10} b$
$\log_{10} (\frac{a}{b}) = 10^{x - y}$

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Answer.

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Remark 5.5.8.

In Activity 5.5.6 and Activity 5.5.7, you explored an example of two common properties of logarithms!

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Definition 5.5.9.

The product property of logarithms states

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\log_{a} (m \cdot n) = \log_{a} m + \log_{a} n .

The quotient property of logarithms states

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\log_{a} (\frac{m}{n}) = \log_{a} m - \log_{a} n .

Notice that for each of these properties, the base has to be the same.

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Activity 5.5.10.

There is still one more property to consider. This activity will investigate the power property. Suppose you are given

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\log (x^{3}) = \log (x \cdot x \cdot x) .

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(a)

By applying Definition 5.5.9, how could you rewrite the right-hand side of the equation?

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$\log (3 x)$
$\log (x) - \log (x) - \log (x)$
$\log (x^{3})$
$\log (x) + \log (x) + \log (x)$

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Answer.

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(b)

By combining "like" terms, you can simplify the right-hand side of the equation further. What equation do you have after simplifying the right-hand side?

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$\log (x^{3}) = \log (3 x)$
$\log (x^{3}) = - 3 \log (x)$
$\log (x^{3}) = \log (x^{3})$
$\log (x^{3}) = 3 \log (x)$

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Answer.

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Definition 5.5.11.

The power property of logarithms states

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\log_{a} (m^{n}) = n \cdot \log_{a} (m) .

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Activity 5.5.12.

Apply Definition 5.5.9 and Definition 5.5.11 to expand the following. (Note: When you are asked to expand logarithmic expressions, your goal is to express a single logarithmic expression into many individual parts or components.)

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(a)

\log_{3} (\frac{6}{19})

$\log_{3} 6 + \log_{3} 19$
$\log_{3} (6 - 19)$
$\log_{3} 6 - \log_{3} 19$
$\log_{3} (6 + 19)$

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Answer.

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(b)

\log ((a \cdot b)^{2})

$2 (\log a + \log b)$
$2 \log a - \log b$
$\log a + 2 \log b$
$2 \log a + 2 \log b$

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Answer.

A and D

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(c)

\ln (\frac{x^{3}}{y})

$3 \ln x + \ln y$
$3 \ln x - \ln y$
$3 (\ln x - \ln y)$
$\ln x^{3} - \ln y$

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Answer.

B. Note that D is also correct, but it is not expanded fully.

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(d)

\log (x \cdot y \cdot z^{3})

$\log x + \log y + 3 \log z$
$\log x + \log y + \log z^{3}$
$\log x - \log y - 3 \log z$
$3 (\log x + y + z)$

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Answer.

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Activity 5.5.13.

Apply Definition 5.5.9 and Definition 5.5.11 to condense into a single logarithm.

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(a)

6 \log_{6} a + 3 \log_{6} b

$6 (\log_{6} a) + 3 (\log_{6} b)$
$\log_{6} a^{6} + \log_{6} b^{3}$
$(\log_{6} a)^{6} + (\log_{6} b)^{3}$
$\log_{6} (a^{6} \cdot b^{3})$

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Answer.

D. B is also correct, but it is not condensed fully.

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(b)

\ln x - 4 \ln y

$\ln x - \ln y^{4}$
$\ln (\frac{x}{y^{4}})$
$\ln {(\frac{x}{y})}^{4}$
$\ln (x \cdot y^{4})$

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Answer.

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(c)

2 (\log (2 x) - \log y)

$\log (\frac{4 x^{2}}{y^{2}})$
$\log (\frac{2 x^{2}}{y^{2}})$
$2 \cdot \log (\frac{2 x}{y})$
$\log (\frac{4 x^{2}}{y})$

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Answer.

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(d)

\log 3 + 2 \log 5

$\log (3 \cdot 5^{2})$
$2 \cdot \log 75$
$\log 75$
$\log 225$

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Answer.

C. Note that A is also correct, but not condensed completely.

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Remark 5.5.14.

You might have noticed that a scientific calculator has only "log" and "ln" buttons (because those are the most common bases we use), but not all logs have base 10 or e as their bases.

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Activity 5.5.15.

Suppose you wanted to find the value of

\log_{5} 3

in your calculator but you do not know how to input a base other than

10

e

(i.e., you only have the "log" and "ln" buttons on your calculator). Let’s explore another helpful tool that can help us find the value of

\log_{5} 3 .

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(a)

Let’s start with the general statement,

\log_{b} a = x .

How can we rewrite this logarithmic equation into an exponential equation?

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Answer.

b^{x} = a

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(b)

Now take the log of both sides of your equation and apply the power property of logarithms to bring the exponent down. What equation do you have now?

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Answer.

x \cdot \log b = \log a

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(c)

Solve for

x .

What does

x

equal?

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Answer.

x = \frac{\log a}{\log b}

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(d)

Recall that when we started, we defined

x = \log_{b} a .

Substitute

\log_{b} a

into your equation you got in part c for

x .

What is the resulting equation?

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Answer.

\log_{b} a = \frac{\log a}{\log b}

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(e)

Apply what you got in part d to find the value of

\log_{5} 3 .

What is the approximate value of

\log_{5} 3 ?

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Answer.

\log_{5} 3

is approximately

0.683

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Remark 5.5.16.

Notice in Activity 5.5.15, we were able to calculate

\log_{5} 3

using logs of base

10 .

You should now be able to find the value of a logarithm of any base!

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Definition 5.5.17.

The change of base formula is used to write a logarithm of a number with a given base as the ratio of two logarithms each with the same base that is different from the base of the original logarithm.

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