Elementary Functions Rules for logarithms Part 3 ......There is the change of base equation: if c is a positive real number then log b x = log c x log c b (4) There are two identities

Elementary FunctionsPart 3, Exponential Functions & Logarithms

Lecture 3.4a, Working With Logarithms

Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 1 / 23

Rules for logarithms

We review the properties of logarithms from the previous lecture. In thatlecture, we developed the following identities.The “Product Property” is an identity involving the logarithm of product:

logb(MN) = logbM + logbN (1)

The “Quotient Property” is an identity involving the log of quotient:

logb(MN ) = logbM − logbN (2)

The “Exponent Property” allows us to rewrite the log of an expressionwith an exponent:

logb(Mc) = c · logbM (3)


Exponential Functions

There is the change of base equation: if c is a positive real number then

logb x = logc xlogc b

(4)

There are two identities that express the inverse relationship betweenexponential and logarithmic functions:

logb(bx) = x. (5)

blogb x = x. (6)


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 inversefunction of the exponential function.


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 basee, then ln e3 = 3 so the answer is

3− ln(x2 + 4) .



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) .



3 By the quotient property,

log100x3y2

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 soour final answer is

2 + 3 log x+ 2 log y − 4 log z .



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 .)


Simplification of logarithms

We can use our six logarithm identities to simplify expressions involvinglogs. 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 inverseproperty to compute logb(

b3

b5) = logb b

3 − logb b5 = 3− 5 = −2.

2 By the exponent property logb((b3)5) = 5 logb b

3. By the first inverseproperty 5 logb b

3 = 5(3) = 15.



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(

13). By the first

inverse property, eln13 = 1

3 .



Sometimes a problem has an answer in a base which is intrinsic to theproblem but it is not a base with which we can easily do computations. Inthat case we need to be prepared to change the base to one with which wecan compute.

Worked problems on changing the base of the logarithm.Use the “change of base” identity to write the following as fractionsinvolving ln(). Use a calculator (or computer software program) toapproximate the answer.

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

Solutions.

1 log2(5) =ln 5ln 2 ≈ 2.3219 .

2 log2(125) =ln 125ln 2 ≈ 6.9658 .

3 log16(17) =ln 17ln 16 ≈ 1.0219 .

4 log(5) = ln 5ln 10 ≈ 0.6990 .

5 log2(1024) =ln 1024ln 2 = 10.0000 .


Logarithms

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

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Elementary FunctionsPart 3, Exponential Functions & Logarithms

Lecture 3.4b, Working With Logarithms, continued

Dr. Ken W. Smith

Sam Houston State University

2013



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



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 andb = ln 2.)

2 By basic properties of exponents, 22x−2 = 22x

22= (14)2

2x. Since

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

is 14e

2(ln 2)x . (Here a = 14 and b = 2 ln 2.)

3 Since 3 = eln 3 then y = 3x = (eln 3)x = e(ln 3)x . (Here a = 1 andb = 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 = 10and b = ln 10.)


Understanding logarithms

Suppose you do not have a calculator. You are asked to compute thelogarithms, base 10, of the first ten positive integers, 1,2,3, . . . , 10. Thatis, 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 aboutlogarithms.

1 Which of these ten logarithms can you compute immediately, withoutany further information?Smith (SHSU) Elementary Functions 2013 16 / 23


2 Suppose I tell you that log(2) = 0.30103. Which of these tenlogarithms can you compute now?

3 Suppose I tell you that log(2) = 0.30103 and log(3) = 0.47712. Nowwhich of these ten logarithms can you compute?

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



Solution.

1 Using properties of logs, without any further information, we knowthat log(1) = 0 and log(10) = 1. So we can fill in the first and lastline:

log(1) 0

log(2)

log(3)

log(4)

log(5)

log(6)

log(7)

log(8)

log(9)

log(10) 1



2 Using log(2) = .30103, we havelog(4) = log(22) = 2 log 2 = 2(0.30103) = 0.60206 andlog(8) = log(23) = 3 log 2 = 3(0.30103) = .90309.

We also havelog(5) = log(102 ) = log(10)− log(2) = 1− .30103 = .69897.



So now we know:log(1) 0

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



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.



So now we know:

log(1) 0

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 aproduct or quotient of powers of 2 and 3.)

Challenge exercise: with this information on log 2 and log 3 how would youcompute log 7.2?


Logarithms

In the next presentation we use our log properties to solve a variety ofequations.

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Elementary Functions Rules for logarithms Part 3 ......There is the change of base equation: if c is a positive real number then log b x = log c x log c b (4) There are two identities