Rules for logarithms We review the properties of logarithms from the - - PowerPoint PPT Presentation

Elementary Functions

Part 3, Exponential Functions & Logarithms Lecture 3.4a, Working With Logarithms

• Dr. Ken W. Smith

Sam Houston State University

2013

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Rules for logarithms

We review the properties of logarithms from the previous lecture. In that lecture, we developed the following identities. The “Product Property” is an identity involving the logarithm of product: logb(MN) = logb M + logb N (1) The “Quotient Property” is an identity involving the log of quotient: logb( M

N ) = logb M − logb N

(2) The “Exponent Property” allows us to rewrite the log of an expression with an exponent: logb(Mc) = c · logb M (3)

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Exponential Functions

There is the change of base equation: if c is a positive real number then logb x = logc x

logc b

(4) There are two identities that express the inverse relationship between exponential and logarithmic functions: logb(bx) = x. (5) blogb x = x. (6)

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Rules for logarithms

The first three equations here are properties of exponents translated into “logarithm language.” The fourth equation allows us to choose the base of our logarithm. The last two equations in the list identify the logarithm as the inverse function of the exponential function.

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Practicing Logarithms

Let’s practice these properties of logarithms on some exercises. Some worked problems. Expand the following expressions.

1 ln e3 x2+4 2 ln (5(x+1))3 (5x−7)2 . 3 log 100x3y2 z4

.

4 log(x3y2z5)12.

Solutions.

1 By the quotient property

ln( e3 x2 + 4) = ln e3 − ln(x2 + 4). By the first inverse property, since ln() stands for the logarithm base e, then ln e3 = 3 so the answer is 3 − ln(x2 + 4) .

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Practicing Logarithms

2 By the quotient property

ln (5(x + 1))3 (5x − 7)2 = ln((5(x + 1))3) − ln((5x − 7)2). By the exponent property ln((5(x + 1))3) − ln((5x − 7)2) = 3 ln(5(x + 1)) − 2 ln(5x − 7). By the product property this is equal to 3(ln 5+ln(x+1))−2 ln(5x−7) = 3 ln 5 + 3 ln(x + 1) − 2 ln(5x − 7) .

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Practicing Logarithms

3 By the quotient property,

log 100x3y2 z4 = log 100x3y2 − log z4. By the product property log 100x3y2 = log 100 + log x3 + log y2. By the exponent property, we can rewrite all the exponents so that log 100+ log x3 + log y2 − log z4 = log 100+ 3 log x + 2 log y − 4 log z. Since log() stands for the logarithm base ten then log 100 = 2 and so

• ur final answer is

2 + 3 log x + 2 log y − 4 log z .

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Practicing Logarithms

4 By the exponent property,

log(x3y2z5)12 = 12 log(x3y2z5) By the product property, 12 log(x3y2z5) = 12(log x3 + log y2 + log z5) and then by the exponent property we have 12(3 log x + 2 log y + 5 log z) . (Also acceptable is 36 log x + 24 log y + 60 log z .)

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Simplification of logarithms

We can use our six logarithm identities to simplify expressions involving

• logs. Here are some worked examples.

More worked problems. Use properties of logarithms to simplify the following expressions.

1 logb( b3 b5 ) 2 logb((b3)5)

Solutions.

1 We could simplify b3 b5 = b−2 and then recognize that logb(b−2) = −2.

Or we could use the quotient property of logs and the first inverse property to compute logb( b3

b5 ) = logb b3 − logb b5 = 3 − 5 = −2. 2 By the exponent property logb((b3)5) = 5 logb b3. By the first inverse

property 5 logb b3 = 5(3) = 15.

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Simplification of logarithms

Use properties of logarithms to simplify the following expressions.

3 loge(e √ 2) 4 10log10(5) 5 e− ln 3,

Solutions.

3 By the first inverse property loge(e √ 2) =

√ 2.

4 By the second inverse property, 10log10(5) = 5. 5 By the exponent property e− ln 3 = eln(3−1) = eln( 1

3 ). By the first

inverse property, eln 1

3 = 1

3.

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Simplification of logarithms

Sometimes a problem has an answer in a base which is intrinsic to the problem but it is not a base with which we can easily do computations. In that case we need to be prepared to change the base to one with which we can compute. Worked problems on changing the base of the logarithm. Use the “change of base” identity to write the following as fractions involving ln(). Use a calculator (or computer software program) to approximate the answer.

1 log2(5). 2 log2(125). 3 log16(17). 4 log(5). 5 log2(1024).

Solutions.

1 log2(5) = ln 5 ln 2 ≈ 2.3219 . 2 log2(125) = ln 125 ln 2 ≈ 6.9658 . 3 log16(17) = ln 17 ln 16 ≈ 1.0219 .

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Logarithms

In the next presentation we continue to practice our logarithm properties. (END)

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Elementary Functions

Part 3, Exponential Functions & Logarithms Lecture 3.4b, Working With Logarithms, continued

• Dr. Ken W. Smith

Sam Houston State University

2013

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Simplification of logarithms

A general exponential function has form y = aebx where a and b are constants and the base of the exponential has been chosen to be e. Occasionally we have an exponential function with a different base and need to change the function into this general form. Here are some examples.

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Simplification of logarithms

Worked problems on general exponential form. Write the following functions in the form y = aebx.

1 y = 2x 2 y = 22x−2 3 y = 3x 4 y = 10x+1

Solutions.

1 Since 2 = eln 2 then y = 2x = (eln 2)x = e(ln 2)x . (Here a = 1 and

b = ln 2.)

2 By basic properties of exponents, 22x−2 = 22x 22 = ( 1 4)22x. Since

2 = eln 2 then 22x = (eln 2)2x = e(ln 2)(2x) = e(2 ln 2) x. So our answer is

1 4e2(ln 2)x . (Here a = 1 4 and b = 2 ln 2.) 3 Since 3 = eln 3 then y = 3x = (eln 3)x = e(ln 3)x . (Here a = 1 and

b = ln 3.)

4 By our basic properties of exponents, 10x+1 = 10x · 10. Since

10 = eln 10 then y = 10x+1 = 10(eln 10)x = 10e(ln 10)x . (Here a = 10

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Understanding logarithms

Suppose you do not have a calculator. You are asked to compute the logarithms, base 10, of the first ten positive integers, 1,2,3, . . . , 10. That is, you are asked to fill out as much of the following table as possible. log(1) log(2) log(3) log(4) log(5) log(6) log(7) log(8) log(9) log(10) Here are a series of questions designed to explore what we know about logarithms.

1 Which of these ten logarithms can you compute immediately, without

any further information?

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Understanding logarithms

2 Suppose I tell you that log(2) = 0.30103. Which of these ten

logarithms can you compute now?

3 Suppose I tell you that log(2) = 0.30103 and log(3) = 0.47712. Now

which of these ten logarithms can you compute?

4 Given the information for log(2) and log(3), fill out as much of the

table as possible.

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Understanding logarithms

Solution.

1 Using properties of logs, without any further information, we know

that log(1) = 0 and log(10) = 1. So we can fill in the first and last line: log(1) log(2) log(3) log(4) log(5) log(6) log(7) log(8) log(9) log(10) 1

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Understanding logarithms

2 Using log(2) = .30103, we have

log(4) = log(22) = 2 log 2 = 2(0.30103) = 0.60206 and log(8) = log(23) = 3 log 2 = 3(0.30103) = .90309. We also have log(5) = log( 10

2 ) = log(10) − log(2) = 1 − .30103 = .69897.

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Understanding logarithms

So now we know: log(1) log(2) 0.30103 log(3) log(4) 0.60206 log(5) 0.69897 log(6) log(7) log(8) 0.90309 log(9) log(10) 1

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Understanding logarithms

3 Using log(2) = .30103 and log(3) = .47712 we have

log(6) = log(3 · 2) = log 2 + log 3 = .30103 + .47712 = .77815. and log(9) = log(32) = 2 log(3) = 2(.47712) = .95424.

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Understanding logarithms

So now we know: log(1) log(2) 0.30103 log(3) 0.47712 log(4) 0.60206 log(5) 0.69897 log(6) 0.77815 log(7) log(8) 0.90309 log(9) 0.95424 log(10) 1 (We cannot obtain the log of 7 this way since we cannot write 7 as a product or quotient of powers of 2 and 3.) Challenge exercise: with this information on log 2 and log 3 how would you compute log 7.2?

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Logarithms

In the next presentation we use our log properties to solve a variety of equations. (END)

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