INTERESTING QUESTIONS & “UNDOING” EXPONENTS: LOGS Hey, hey logarithm…When we get the blues.

INTERESTING QUESTIONS &“UNDOING” EXPONENTS: LOGS

Hey, hey logarithm…When we get the blues

Recap

Last week we looked at RATIONAL exponents and saw that

A square root is the same as an exponent of ½ A square root is the same as an exponent of ½

A cubed root is the exponent 1/3 A cubed root is the exponent 1/3

To evaluate powers with rational exponents, we “rip the exponent apart”. 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 radioactive materials will decay in an exponential fashion (half-life)

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

InterestInterest

In general, the compound interest is nt

n

rPA

1

Where A is the amount in the account at time, tP is the principle (initial) amount r is the decimal value of the interest rate n is how many times per year the interest is compounded.

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

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

InterestInterest

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

A new type of question?

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 get

Solving for the Exponent

t120035.12We’re totally stuck! We presently have no way of solving for an exponent unless we can get common bases.

We’re totally stuck! We presently have no way of solving for an exponent unless we can get common bases.

So mathematicians invented logarithms.

823 can also be written as

38log2

basebase

exponentexponent

argumentargument

argumentargument

basebase

exponentexponent

We read this “log to base 2 of 8 equals 3.We read this “log to base 2 of 8 equals 3.

Going from one form to another

932

Write the following 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 a ‘10’ we do not need to write it. This is because base 10 is what most calculators deal with.

When the base is a ‘10’ we do not need to write it. This is because base 10 is what most calculators deal with.

Going from one form to another

Write the following in exponential form.

5.09log81

3

1

5

1log125

3

24log8

2

150log50

29

4log

23

201.0log

981 5.0

5

1125 3

1

4832

505021

9

4

2

32

01.010 2

cab log

abc

Evaluating Logs

Evaluate: a) 32log2 This asks “2 to the what gives 32?”

or 322 x We know this is 5.

532log2

b) 64log4 This asks “4 to the what gives 64?”

or 644 x

We know this is 3.364log4

c) 32log41 This asks “1/4 to the what

gives 32?”32

4

1

x

We can get common bases

52 22 x

5.2

52

x

x

By changing forms we can evaluate log expressions.

d) log 100 This asks “10 to the what gives 100”?

We know this is 2.

5.232log41

Solving log equations: common bases

To continue getting used to logs, we’ll look at these questions.

Solve for x.

3log2 x

x 32

We’re stuck in log form so go to exponential form

We can solve this: x =1/8

b) x9log3


a)

22

22

1

33

33

93

x

x

x

4

22

x

x

c) 5

416log x


165

4

x To solve for the base, we can undo the exponent by raising both sides to the 5/4.

4

54

5

5

4

16

x

5

4

1

4

5

16

16

x

x 32

2 5

x

x

d) x3.1log


But wait… the base here is

10

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

log1.3=0.114

Solving log equations: common bases

a.

b.

c.

d.

e.

f.

g.

h.

Try some…

1) Solve each equation

x 27log32log 32

100loglog2 x

3

2

9

4log x

5010 x

x25

134 log25log81log4

1log

)4(log5log 22 x

x632 log27

1log8log

x 27log32log 32a.

x 27log5 3

y27log3

273 y

2

3

33

33

2

3

2

13

y

y

y

2

72

3

2

102

35

x

x

x

x272 log16log3

1

a.

b.

c.

d.

e.

f.

g.

h.

Try some…


x 27log32log 32

100loglog2 x

3

2

9

4log x

5010 x

x25

134 log25log81log4

1log

)4(log5log 22 x

x632 log27

1log8log

b.

2log2 x

4

22

x

x

100loglog2 x

x272 log16log3

1

a.

b.

c.

d.

e.

f.

g.

h.

Try some…


x 27log32log 32

100loglog2 x

3

2

9

4log x

5010 x

x25

134 log25log81log4

1log

)4(log5log 22 x

x632 log27

1log8log

c.

38log2

x632 log27

1log8log

327

1log3

x

x

6

6

log0

log)3(3

1

60

x

x

x272 log16log3

1

a.

b.

c.

d.

e.

f.

g.

h.

Try some…


x 27log32log 32

100loglog2 x

3

2

9

4log x

5010 x

x25

134 log25log81log4

1log

)4(log5log 22 x

x632 log27

1log8log

d.

14

1log4

481log3

2

21

x

x

x25

134 log25log81log4

1log

225log5

1

x

x

2

2

log1

log)2(41

x272 log16log3

1

a.

b.

c.

d.

e.

f.

g.

h.

Try some…


x 27log32log 32

100loglog2 x

3

2

9

4log x

5010 x

x25

134 log25log81log4

1log

)4(log5log 22 x

x632 log27

1log8log

e.

5010 x

Since I can’t get common bases, I’m stuck in exponential form. So I go to log form.

x50log10

x50log My calculator can find this.

My calculator can find this.

70.1x

x272 log16log3

1

a.

b.

c.

d.

e.

f.

g.

h.

Try some…


x 27log32log 32

100loglog2 x

3

2

9

4log x

x25

134 log25log81log4

1log

)4(log5log 22 x

x632 log27

1log8log

f.

5010 x

I’m stuck in log form. So I go to exponential form.

9

43

2

x

3

2

9

4log x

To solve for the base, I undo the exponent. I raise both sides to the -3/2

2

32

3

3

2

9

4

x

2

3

4

9

x

The negative in the exponent means I “flip the base”

3

2

1

4

9

x 3

2

3

x 8

27x

x272 log16log3

1

a.

b.

c.

d.

e.

f.

g.

h.

Try some…


x 27log32log 32

100loglog2 x

3

2

9

4log x

x25

134 log25log81log4

1log

)4(log5log 22 x

x632 log27

1log8log

x272 log16log3

1

g.

5010 x

x272 log16log3

1

x27log)4(3

1

416log2

x27log3

4



x3

4

27 Rip the exponent apartRip the exponent apart

x

4

3

1

27 x43 x81

a.

b.

c.

d.

e.

f.

g.

h.

Try some…


x 27log32log 32

100loglog2 x

3

2

9

4log x

x25

134 log25log81log4

1log

)4(log5log 22 x

x632 log27

1log8log

x272 log16log3

1

h.

5010 x

)4(log5log 22 x

I don’t know what is.

5log2

But I do 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

So What?We can now write an equation where we solve for the exponent but how do we evaluate the exponent if we do not recognize the log or if we cannot get common bases?

All of these started with the equation

t120035.12 Are we any closer?

Well we can write it in a log form

t122log 0035.1 But that’s as far as we can get.

Notice that theis equal to an exponent. It must also be an exponent.

2log 0035.1

How do the laws of exponents relate to logs?

The Laws of Logs

Remember the laws of exponents:

When multiplying powers with the same base, we keep the base and add the exponents.

When multiplying powers with the same base, we keep the base and add the exponents.

cbcb aaa

xab yac

cbayx ))((

Let

Go to log form cbxya log

yxxy aaa logloglog When adding logs with the same base, we keep the log and base and multiply the arguments.

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

If then xb alog xab If then yc alog yac

The Laws of Logs

Remember the laws of exponents: cb

c

b

aa

a

xab yac

cbay

x

Let

Go to log form

cby

xa

log

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

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

If then xb alog xab If then yc alog yac

yxy

xaaa logloglog

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

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

Laws of Exponents

Practice with the first two laws. Solve for x.

25log2log8log4log.2# 5 xxx

22log)84(log xx

22

32log

x

216log xI’m stuck in log form. So I go to exponential form.


4

162

x

x

x233 log7log189log.1#

x23 log7

189log

x23 log27log

x2log3

8

23

x

x

The 3rd Law of Logs

When we have a power of a power, we keep the base and multiply the exponents.

When we have a power of a power, we keep the base and multiply the exponents. Le

t ant

If then at nlog ant

Raise both sides to the exponent “b”

bbt an btb an Go to log form

bn abt log bnn aab loglog

The “down in front” rule!

Ex. Evaluate

27log3

21

3 27log

27log2

13

2

3

And finally…

Are we any closer to solving the original question?

t120035.12 Let’s take the log of both sides…

t120035.1log2log Now the “down in front” rule

0035.1log122log t Divide by 12log1.0035

t0035.1log12

2logAnd my calculator can do this

53.16t

A shortcut to the calculator rule

So we have seen that t120035.12 can be written as

t120035.1log

2log

So we do not need to take the log of both sides. We can go to log form

t122log 0035.1 And then write t120035.1log

2log

Remember that the base is on the bottom!

Lots o’ Logs

Solve for x.

1003 9 x Since I can’t get common bases, I’m stuck in exponential form. So I go to log form.

9100log3 x

93log

100logx

919.4 x 81.4x

Lots o’ Logs

Solve for x.

8047 x

x74log

80log

Since I can’t get common bases, I’m stuck in exponential form. So I go to log form.

x780log4

84.3

716.3

716.3

x

x

x

INTERESTING QUESTIONS & “UNDOING” EXPONENTS: LOGS Hey, hey logarithm…When we get the blues.

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