47: More Logarithms and Indices © Christine Crisp “Teach A Level Maths” Vol. 1: AS Core...

47: More Logarithms 47: More Logarithms and Indicesand Indices

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 1: AS Core Vol. 1: AS Core ModulesModules

More Logarithms and Indices

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Module C2


We need to be able to change between index forms for numbers and log forms.

bxba ax log

We use

We’ll also develop some more laws of logs.


6443

e.g. Write the following in a form using logarithms:

279 23

(a)

(b)

6443 (a) 64log3 4

27log923

e.g. Write the following without using logarithms: 481log 3

21

4 2log (a)

(b)

Solution: (a)

481log 3 8134

24 21

279 23

(b)

21

4 2log (b)

The index, 3, is the log of 64 and 4 is the base.

Solution: bxba ax log


932

1. Write the following in a form using logarithms:

416 21

(a)

(b)

Solution: 1(a)

932 9log2 3

4log1621

2. Write the following without using logarithms:

3125log5 23

4 8log (a)

(b)

12553

84 23

3125log5 2(a)

Exercises

416 21

(b)

23

4 8log (b)


Some logs can be simplified.

Simplifying Logs

9log 3e.g. 1 Simplify

239

This log can be simplified because we can write 9 in index form using the base 3.

So, 9log 32

3 3log

The base, 3, is now the same as the base of the log

2 since a log is an index!

We are not solving an equation!

ka ka logIn

general,


Simplifying Logs

42 2log

16log 2e.g. 2. Simplify (a) (b)

3

1log9

21

9

1log9

21

9log9

21

(b)

3

1log9

16log 2Solution: (a)

4

9

1log9


100log10

1. Simplify the following log expressions:(a

)(b)

210 10log

5log5 81

2log(c)

(d)

64log4

Solution (a) 100log10

2

(b)

64log4

)4(log 34

3

(c)

5log521

5log5

21

81

2log(d)

322

1log

32 2log 3

Exercises


There are 2 special cases we can get directly from the definition of a log.

Let x = 0,

2 useful results

bxba ax log

ba 0 1b

So, 1log010aa

By the law of indices,

for all values of the base01log a



Let b = a,

x = 1

ax alogaa x Then

2 useful results

bxba ax log


aa x


Let b = a,

x = 1

1Then

aalog

So, 1log aa

2 useful results

bxba ax log


SUMMARY

bxba ax log

1log aa

Three Laws of Logarithms

The Definition of a Logarithm

01log a

ka ka log


1. Simplify the following: 4log4

1910log10b

a alog

(a)

(c)

(b) 1log 2

(d)

(a) 1Ans:

(b) 0

(c) 19 (d) b

Exercises


The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.


6443

e.g. Write the following in a form using logarithms:

279 23

(a)

(b)

6443 (a) 64log3 4

27log923

e.g. Write the following without using logarithms: 481log 3

21

4 2log (a)

(b)

Solution: (a)

481log 3 8134

24 21

279 23

(b)

21

4 2log (b)

Solution: bxba ax log


Simplifying Logs

42 2log

16log 2e.g. 2. Simplify (a) (b)

3

1log9

21

9

1log9

21

9log9

21

(b)

3

1log9

16log 2Solution: (a)

4

9

1log9



Let x = 0,

2 useful results

bxba ax log

ba 0 1b

So, 1log010aa

By the law of indices,

for all values of the base01log a


aa x Let b = a,

x = 1

1Then

aalog

So, 1log aa

bxba ax log


bxba ax log

1log aa

Three Laws of Logarithms

The Definition of a Logarithm

01log a

ka ka log

SUMMARY

47: More Logarithms and Indices © Christine Crisp “Teach A Level Maths” Vol. 1: AS Core...

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