Logarithms: “undoing” exponents. Recap Last week we looked at RATIONAL exponents and saw that A square root is the same as an exponent of ½ A cubed root

Logarithms:“undoing” exponents

RecapLast week we looked at RATIONAL exponents and saw that

A square root is the same as an exponent of ½

A cubed root is the exponent 1/3

To evaluate powers with rational exponents, we “rip the exponent apart”.

We also saw that radioactive materials will decay in an exponential fashion (half-life)

We also saw that compound interest can be modeled using exponential equations

nt

n

iPA

1

Where

A is the $ amount in the account at time t (years)P is the principle (initial) $ amount (when t = 0)i is the decimal value of the annual interest raten is how many times per year the interest is compoundedt is the number of years

Look for terms like:daily (n = 365),weekly (n = 52)quarterly (n = 4)semi-annually (n = 2) monthly (n = 12)

Compound Interest Formula

Ex 1. A credit card charges 24.2% interest per year compounded monthly. There are $900 worth of purchases made on the card. Calculate the amount owing after 18 months. (Assume that no payments were made.)

)(12

12

242.01900

t

A

)5.1(12020166667.1900A

21.1289$A

nt

n

iPA

1

More Compound Interest Examples

P = 900i = 24.2% = 0.242n = 12t = 1.5y = ?

Ex 2. A bank account earns interest compounded monthly at an annual rate at 4.2%. Initially the investment was $400. When does it double in value?

So this questions seems to be like all the others…

t

y12

12

042.01400

t120035.1400800

t120035.12 And now we geta common base……except we can’t.

nt

n

iPA

1P = $400

i = 4.2% = 0.042n = 12t = ?y = $800

More Compound Interest Examples

…maybe by the end of class…

Logarithmic FormWe`re totally stuck. So far we can only solve for an exponent when we can achieve common bases. Good thing we have logarithms to help.

823 can also be written as 38log2

t120035.12

This form is helpful because the exponent value is no longer locked in the exponent position, in fact it is isolated.

That is “log base 2 of 8 equals 3”

exponent exponent

basebase argument argument

Going from exponential to logarithmic form

932

Write the following exponential equations in logarithmic form.

81643

25

1125 3

2

130

1000103

63621

29log3

2

16log36

4

38log16

3

2

25

1log125

01log3

31000log

When the base is 10, we usually leave it out… take a look at your calculator…

5.09log81 3

1

5

1log125

3

24log8

2

150log50 2

9

4log

23 201.0log

981 5.0 5

1125 3

1

4832

505021

9

4

2

32

01.010 2

cab log

abc

Going from logarithmic to exponential form

Write the following logarithmic equations in exponential form.

Evaluating Logs

Examples32log2

This asks “2 to the what gives 32?”

or, solve for x: 322 x

We know this is 5, so: 532log2

b) Evaluate: 64log4

This asks “4 to the what gives 64?”

or, solve for x: 644 x

We know this is 3, so: 364log4

By changing forms we can evaluate log expressions.

a) Evaluate:

364log4

532log2

Evaluating Logs

This asks “1/4 to the what gives 32?”

or, solve for x:

We can get common bases :

d) Evaluate: 100log or, solve for x: 10010 x

We know this is 2, so:

c) Evaluate: 32log41 32

4

1

x

52 22 x

5.2

52

x

x

5.232log41

2100log

5.232log41

2100log

Solving log equationsSolving logarithmic equations takes some instinct, which only comes from practice, but to help you get you started, here is a flowchart with some possibly useful steps.

Simplify:evaluate any

complete log or exponential expressions

Isolate the unknown:

If the unknown is in the…

…argument: change it to exponential form

…exponent: in exponential form get common bases if possible, or change to logarithmic form to solve, or take the log of both sides and apply log

rules

…base: write in exponential form then remove the exponent by raising each

side to the opposite exponent

Solving log equations - Examples

3log2 x

x 32

b) Solve for x. x9log

3

22

221

33

33

x

x

4

22

x

x

a) Solve for x.

8

1

2

13

x

x

93 x

Unknown in the argument

Unknown in the exponent

c) 5

416log x

165

4

x

454

5

54

16

x

5

41

4

5

16

16

x

x

32

2 5

x

x

Solving log equations

Unknown in the base

d)

But wait… the base here is... 10

Since the calculator uses base 10, just type this in and get…

x3.1log

114.03.1log

Solving log equations

Unknown in the argument

a.

b.

c.

d.

e.

f.

g.

h.

Solve each equation for x

x 27log32log 32

100loglog2 x

3

2

9

4log x

5010 x

x25134 log25log81log

4

1log

)4(log5log 22 x

x632 log27

1log8log

x272 log16log3

1

Solving log equations - Practice

Solutions

x 27log32log a. 32

y27log3

273 y

2

3

33

33

2

3

2

13

y

y

y

2

72

3

2

102

35

x

x

x

532log2

2

327log3

Evaluations


Solutions

100loglog b. 2 x x632 log27

1log8log c.

2log2 x

4

22

x

x 38log2

327

1log3

x

x

6

6

log0

log)3(3

1

60

x

x

2100log

Solving log equations - PracticeEvaluations

Solutions

x25134 log25log81log

4

1log d.

14

1log4

481log3

2

21

x

x225log

51

x

x

2

2

log1

log)2(41


Evaluations

Solutions

5010 e. x

x50log10

x50log70.1x


Solutions

3

2

9

4log f. x

9

432

x

2

32

3

3

2

9

4

x

23

4

9

x

3

21

4

9

x

3

2

3

x

8

27x


Solutions

x272 log16log3

1 g.

416log2 x27log)4(

3

1

x27log3

4

x34

27

x

4

31

27

x43

x81


Evaluations

Solutions

We don’t know what is.)4(log5log h. 22 x 5log2

But notice that there’s a common base on both sides of the equation.

Since the bases are equal, the ARGUMENTS must be equal.

9

)4(5

x

x


Remember the laws of exponents:

cbcb aaa xab yac Let and

cbxya log

yxxy aaa logloglog

yc alogxb alog

cbaxy

So and

The first law of logarithms

When multiplying powers with the same base, we

keep the base and add the exponents.

When adding logs with the same base, we keep the log and base and multiply

the arguments

xyyx aaa logloglog

First law of logarithms:

xyyx aaa logloglog


xab yac

Let and

yc alogxb alogSo and

The second law of logarithms

Second law of logarithms:

yxy

xaaa logloglog

cbc

b

aa

a

cbay

x

cby

xa

log

When dividing powers with the same base, we keep the base and subtract the

exponents.

When subtracting logs with the same base, we keep the log and the base and

divide the exponents.

y

xyx aaa logloglog

Practice with the first two logarithm laws. Solve for x.

25log2log8log4log.2# 5 xxx

22log)84(log xx

22

32log

x

216log x

4

162

x

x

x233 log7log189log.1#

x23 log7

189log

x23 log27log

x2log3

8

23

x

x

Laws of Logarithms - Practice

22log)32(log xx


ant Let

at nlogSo

The third law of logarithms

Third law of logarithms:The “down in front” rule When we have a power of

a power, we keep the base and multiply the exponents.

If the argument of a logarithm is a power, the exponent can be moved “down in front”.

btbt nn

bn abt log

bnn aab loglog

btb na

aba nb

n loglog

Practice with the third logarithm law.

27log3 2

1

3 27log

27log2

13

32

1


2

3

Evaluate. Solve.

2.04 32 x

2.0log4log 32 x

05.1

16.33

216.13

16.132

x

x

x

x

4log

2.0log32 x

2.0log4log32 x

yba log

This third law of logs is the key to evaluating logs with bases other than 10 on your calculator!

take the log of both sides

apply the “down in front” rule.

bay bay loglog

change to exponential form

bay loglog

a

by

log

log

a

bba log

loglog

balog yba logLet

Shortcut for evaluating logs

yba logremember that

Logarithm Shortcut 1:To evaluate logab on your calculator, divide logb by loga.

a

bba log

loglog Remember,

the base is on the bottom

abx

This third law of logs is the key to solving exponential equations when common bases can’t be achieved!

abx loglog

abx loglog

b

ax

log

log

take the log of both sides

apply the “down in front” rule.

Shortcut for solving for exponents

Logarithm Shortcut 2:To solve for an exponent, divide the log of the argument by the log of the base.

b

axabx

log

log then , if

Remember,the base is on the

bottom

Practice with the third logarithm law.

27log3 2

1

3 27log

27log2

13

32

1


2

3

Evaluate. Solve.

2.04 32 x

2.0log4log 32 x

05.1

16.33

216.13

16.132

x

x

x

x

4log

2.0log32 x

2.0log4log32 x

Evaluate again.

3log

27log27log3

5.1

Solve again.

2.04 32 x

4log

2.0log32 x

05.1

16.33

216.13

16.132

x

x

x

x

Solve for x.

10093 41 x

10033421 x

10033 81 x

1003 9 x

3log

100log9 x

81.4

19.49

x

x

402 74 x

742log

40log x

743219.5 x

08.3

43219.12

x

x

Laws of Logarithms – Practice

617log2 x

617log1

2 x

617log1

2 x

x617log2

x62log

17log

68.0

609.4

x

x

logb(ba) = a

Another shortcut forevaluating certain logarithms

3

5log125log 355

125log5

Examples

128

1log2

7

2log128

1log 7

22

81log31

4

3

1log81log

4

31

31

Challenge: Can you prove why this shortcut is true? logb(ba) = a

Ex 2. A bank account earns interest compounded monthly at an annual rate at 4.2%. Initially the investment was $400. When does it double in value?

So this questions seems to be like all the others…

t

y12

12

042.01400

t120035.1400800

t120035.12 We can’t get common bases… but…..

nt

n

iPA

1

P = $400i = 4.2% = 0.042n = 12t = ? yearsy = $800

And back to the beginning…We can now solve this question!

0035.1log

2log12 t

t120035.12

53.16t36.19812 t

Documents

Logarithms: “undoing” exponents. Recap Last week we looked at RATIONAL exponents and saw that A square root is the same as an exponent of ½ A cubed root