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### Transcript of Indices and Logarithms

• 910YEARS

NUMBER AND ALGEBRA Module 31

A guide for teachers - Years 910 June 2011

• Peter Brown

Michael Evans

David Hunt

Janine McIntosh

Bill Pender

Jacqui Ramagge

910YEARS

INDICES AND LOGARITHMS

Indices aLogrithmndehmo(NiubA(hmoagiogil(naa:NiMIctul3i1 a)d(mi NUMBER AND ALGEBRA Module 31

A guide for teachers - Years 910 June 2011

• LOGARITHMS

{4} A guide for teachers

• 4

5

EXERCISE 1

'

The Zero Index

EXAMPLE

• Negative Exponents

If we examine the pattern forme d when we take decreasing powers of 2, we see

24 = 16, 23 = 8, 2 2 = 2, 21 = 2, 20 = 1, 21 = ?, 22 = ?

At each step as we decrease the index, the number is halved. Thus it is sensible to dene

21 = 12.

Furthermore , con tinuing t he pattern, we dene

22 = 14 = 122, 2

3 =

18 =

123, and so on.

These denitions ar e consistent with the index laws.

For example, 23

24 = 23 4

= 2 1 . But clearly, 23

24 = 12.

Similarly, 22

24 = 22 4

= 2 2 . but clearly, 22

24 = 14 =

122.

We can con rm o ur intuition by c onsidering and .

In general, for any non-zero number , and positive integer , we dene

a1 = 1a and an

= 1an.

Note that all the earlier index laws also hold f or negative indices.

EXAMPLE

Simplify

a 82 b 233

c (a2)3b3

ab2

2

SOLUTIONa 82 = 164 b

23

3=

32

3 =

278

c ab3

ab2

2 =

a6

b3 a2

b 4 = a6

b3 b 4a2 = a

4 b 7 = 1a4b7

Notice that

ab

1 =

ba

When asked to simplify an expression involving indices, we generally express our answer using positive indices.

EXERCISE 2

Simplify x1

+ y1x2 y2 .

2

• fi

fl

'

• Note that we co uld also have expressed this as the cube root of the square of 8, w hich is, of course, also equal to 4, that is:

823 = 823 = 643 = 4 or,

823 = ( 83 )2 = 22 = 4.

In general, if is a positive number and are positive integers, we dene

apq = a

1q

p

or apq

.

In words, we take the qth root of a and raise it to the p ower p.

EXAMPLEFind a 16

34 b

49

12

SOLUTIONa 16

34 = ( 164 )3 = 23 = 8 b 49

12

=

4 9

3

=

23

3

= 827.

Negative fr acti onal ind ices

Finally, we can extend the indices to include negative rationals. For example,

813 = 8

13

1

= 1

813 =

12.

So that

apq =

1a

pq.

EXERCISE 3 Show that: 32

25 =

14 ,

1681

14

=

32, 2x

25

5

=

32x2 .

We have now de ned ax for any positive real number a and any rational number x. It remains to check that the index laws also hold i n this more general case. We will not go through the details. The following example outli nes how this m ight be done in one particular case.

EXERCISE 4This exercise asks you to give a proo f that the rs t index law hol ds for negative integer exponents and also for fractional exponents.

a By writing a pa q as 1ap

1aq , show that a pa q = a p q as, where p, q are positive integers.

b By writing as show that a1n

a1m as a

mmn

an

mn = a

mnm

amn

n

, show that a1n

a1m = a

1n +

1m

.

It is possible to give similar proof s that the ot her index laws also hold f or negat ive integer and rational exponents.

• SCIENTIFIC NOTATION

Scientic notation , or standard form, is a convenient way to represent very large or very small numbers. It allow s the numbers to be easily recorded and read.

The star Sirius is approximately 75 684 000 00 0 000 km from the sun. We can represent this numb er more c ompactly by moving the decimal point to just after the rst non-zero digit and multiplying by an appropriate power o f 10 to rec over the original number . Thus

75 684 000 000 000 = 7.5684 10 13.

If we mov e the decim al point 13 places to the right, inserting the necessary zeroes, we arrive back at the number we started with.

We can similarly deal with very small numb ers using negative indices. For example, an Angstrom () is a unit of length equal to 0.00 0 000 000 1 m, whi ch is the approximate diameter o f a small atom. We place the decimal point just after the rs t non-zer o digit and mul tiply by the appropriate power o f ten. Thus,

0.000 000 0 00 1 = 1 10 10 . Hence, fo r example, the diameter o f a uranium atom is 0.000 000 0 00 38 m which we m ay write as 3.8 10 10 m or 3.8 .

The index laws may be used to perform op erations on numb ers written in scientic notati on.

EXAMPLESimplify (3.14 10 2 )3 (7.1 10 8 ) giving your answer correct to o ne decimal p lace.

SOLUTION(3.14 10 2 )3 (7.1 10 8 ) = (3.143 7 .1) 10 2 4.36044 102 436.0 correct to 1 decimal place. In this case, we could leave this as the answer, or, if required, write is as 4.36 10 2.

Signicant gur es in scientic not ation

Scientists and engineers routine ly employ scientic notation to r epresent large and small numbers. Since all measurements are approximat ions anyway, they generally report the numbers rounded to a given number of signicant gures. Thus, a number such as 2.1789 10 7 could be writt en as approximately 2.18 10 7. This is the same as rounding the number 21 789 000 to 21 800 000 , that is, correct to three signica nt gures.

signi ca nt gures . For example, 3.1 has 2 signi cant gures, 3.14 has 3 signica nt gures and so on. To round a number to a required number of signicant gures, r st write the number in scientic notation and identify the last signi cant digit required. Then leave the digit alone if the next digit is 0, 1, 2 ,3 or 4 (in this case the original numb er is rounded down) and increase the last digit by one if the next digit is 5, 6, 7, 8 or 9 (in this case the original numb er is rounded u p.)

• EXERCISE 6

• EXERCISE 7

• exponential growth.

• EXERCISE 8

• EXAMPLE

9

• Logarithms to the base 10

• Law 4

Law 5

a b c

a b

c

EXERCISE 10

a

b

c

• Change of base

EXAMPLE

The Logarithm graph

• EXERCISE 11

• EXERCISE 12

PGP

20

• EXAMPLE

• fl

'

• Log plots

EXERCISE 2

EXERCISE 4

EXERCISE 5

EXERCISE 6

EXERCISE 7

EXERCISE 8

• c

i ii

iii

a

b

c

d

e

f

a

b

c

• EXERCISE 12

EXERCISE 13

• {27}The Improving Mathematics Education in Schools (TIMES) Project