MATHEMATICS Learnerâ€™s Study and Revision Guide for Grade 12 LOGARITHMS

Revision notes and exercises by Roseinnes Phahle

Preparation for the Mathematics examination brought to you by Kagiso Trust

Contents Unit 3 Logarithms: meaning and definition

3

Numbers that are not powers of 10

4

Relationship between logs and index laws

4

Laws of logarithms

5

More examples

5

Exerc ise 3.1

8

Applications of logarithms: Exercise 3.2

9

Answers

10

About this guide 1. This guide gives you a complete revision of logarithms: their meaning, how to evaluate them using an electronic calculator, simplifying expressions involving logarithms, solving equations requiring the use of logarithms and real life examples of the applications of logarithms. 2. You can work out your solutions to the examples in the blank spaces provided. 3. Donâ€™t work alone. Work through this guide in a team with your classmates.

Logarithms

REVISION UNIT 3: LOGARITHMS The logarithm of a number is the index (power) to which a base must be raised to give that number. As an example, 16 = 2 4 , this means that 4 is the logarithm of 16 to base 2. In practice, the name Logarithm is abbreviated to Log:

16 = 2 4 which is in index form â†” Log 2 16 = 4 which is in logarithmic form Here are further examples in which the interchange between index and logarithmic forms is shown:

1000 = 10 3

Log10 1000 = 3

100 = 10 2

Log10 100 = 2

10 = 101

Log10 10 = 1

When the base is 10 we need not show it in writing so that we simply write:

Log1000 = 3 Log100 = 2 Log10 = 1 Evaluating logs using a calculator You can verify all of the above results using the Log key on a calculator. For example, to find the log of 1000 press the following keys in the sequence shown:

What do you get?

log 1000 ) = The log key on the calculator works only when the base is 10.

We need not use a calculator to find the logs of numbers that are powers of 10 as in the above examples. The logs of such numbers are simply the powers to which these numbers are raised when written to base 10. Thus log 1 000 000 = 6 because 1 000 000 = 10 6 . Check this on your calculator by keying in log 1 000 000 = 3

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Numbers that are not a power of 10 What happens though when we have to find the log of a number that is not a simple power of 10? In this case you have no choice but to use a calculator. Example:

log 5342 = 3,7277 to 4 places of decimals. Check this on your calculator.

Example:

Thus in index form, the above is 5342 = 10 3, 7277

The antilog You can check the last example by using what is known as the anti-log key given by the sequence: SHIFT

10 x Example:

Evaluate the anti-log of 3,7277

Solution:

On your calculator enter the following keys in the sequence shown: SHIFT

10 x 3,7277 = What do you get? Relationship between logs and index laws As we have seen, a logarithm is simple the power or index to which a number is raised given a base. There is therefore a relationship between the laws governing logarithms and indices. Example:

368 × 0,583 = 10 2,5658 × 10 −0, 2343 writing each number in index form = 10 2,5658+ (−0, 2343 ) adding indices because we are multiplying = 10 2,5658−0, 2343 = 10 2,3315

What this implies is that

log(368 × 0,583) = log 368 + log 0,583 log(214,544 ) = 2,5658 + (− 0,2343) 2,3315 = 2,3315

Logarithms

To finish the multiplication, do the antilog of 2,3315 SHIFT

10 x 2,3315 = What do you get? First law of logarithms From the foregoing, we can deduce that log AB = log A + log B Second law of logarithms In a similar way we can illustrate the second law which is

log

A = log A - log B B

Third law of logarithms The third law is logA x = xlogA

log12 3 = 3 log 12

Example:

= 3× 1,07918 = 3,2375 Check this answer by working out log12 3 = log 12 × 12 × 12 = log 1728 = What do you get? More examples 1. Log ABCD = log A + log B Log C + log D 2.

log

AB = log A + log B - log C C

3.

log

1 1 = log 3 = log 10 −3 = −3 log 10 = −3 × 1 = −3 1000 10

5

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RECALL: Only when the base is 10 that we need not write it in. Otherwise when the base is not 10 we can convert to base 10 by applying the change of base formula which is at the top of the next page. What you should know Convert from exponential (index) to logarithmic form

Convert from logarithmic to exponential (index) form

Example

Work out the solutions in the boxes below

5 3 = 125

Log 1 81 = âˆ’4 3

log A A = 1

Simplify log 8 8 + 2log 3 3

log A 1 = 0

Simplify 5log16 1

log n A p = plog n A

Simplify log 5 25

Simplify logarithms

log10 100 + log 6 216 + log 3

1 27

Logarithms

What you should know Change of base:

logA log B A = logB

log B A =

1 log A B

log 1 A = −log B A B

Example

Work out the solutions in the boxes below

Use a calculator to evaluate

log 9 17

Write in an alternative form:

log 2 9

Write in an alternative form and simplify: log 1 32 2

Solve equations

Solve inequalities

Solve 6 x = 12 (Take logs on both sides and use a calculator)

log 1 x ≥ 4 3

(Note: logx is only defined for x > 0)

Inverse function NOTE: The inverse function is covered in UNIT 6 but can you at this stage answer this question? NOTE: The three forms in which the inverse function can be written.

Write the inverse of

f ( x ) = log a x in the forms:

y= 1.

f

−1

(x ) =

−1

:x→

and

f

7

Preparation for the Mathematics examination brought to you by Kagiso Trust

The following exercise involves expressions and equations requiring a knowledge of logarithmic laws such as you have practiced above. EXERCISE 3.1 QUESTION 1 Simplify the following expressions: 1.1

log 3 3 − log 3 81 + log 3 1

1.2

1 log 1 243 + log 2 64 4 3

1.3

log k k 5 − 2log k k

1.4

log x

1 1 + log 1 x + log m m + log x x + log 1 x x x x

QUESTION 2

QUESTION 3

Solve the following equations:

Solve the following to 4 decimal places:

2.1

x = log 2 128

3.1

2x = 7

2.2

log 3 x = 4

3.2

43x = 9

2.3

1 log 2 = 3 x 64

3.3

5 2 x +1 = 17

2.4

log x 8 = 2

3.4

2.(3 x ) = 2,38

2.5

logx + log5 = log( x + 16 )

3.5

3.(2 2 x ) = 0,5

2.6

log 3 x + log 3 (x + 6 ) = 3

3.6

x = log 5 8

2.7

log 2 x( x + 2 ) = 3

3.7

25(1 + 0,025) x = 100 x

2.8

log( x + 7 ) = 2log( x + 1)

3.8

1 1 3 = 3 3 3

2.9

log(2 x + 1) − log(x − 1) = 1

3.9

2 x .7 x = log14

Logarithms

APPLICATIONS OF LOGARITHMS EXERCISE 3.2 QUESTION 1 The formula for compound interest is A = P(1 + i )n , where A is the amount after n interest periods, P the initial amount or principal, and i the interest per period. How many years will it take for money to treble when it is compounded annually at 5,5% per year? Give your answer in years and correct to the nearest month. (Hint: Let A = 3P .) QUESTION 2 In the above problem after how many years will it take for the money to treble if it is compounded monthly at the same rate of interest per year? Give your answer in years and correct to the nearest month. (Hint: i = interest rate divided by number of months in the year and n will then be number of months.) QUESTION 3 The animal population of a certain game reserve is estimated to be falling at a yearly rate of 9% of its total population at the beginning of each year. Determine after how long it will take for the population to decrease to 1500 if the population is estimated at 35 000 at the beginning of a particular year. Give your answer in years and correct to the nearest month. QUESTION 4 A steel bar whose temperature is To =1500 C will obviously undergo a fall in temperature if it is immersed in a liquid whose temperature is Tl = 18 C . Given that the temperature T of a hot object placed into a cooling liquid is calculated by using the formula T = Tl + (To − Tl )10 −0, 08t where t is the time in minutes, how long will it take for the bar to cool to 35 C ? Give your answer in minutes and correct to the nearest second. QUESTION 5 In a chrome electroplating process, the mass m in grams of the chrome plating increases according to the formula m = 200 − 2 t 2 , where t is the time in minutes. How many minutes does it take to form 120 g of the plating? Give your answer in minutes and correct to the nearest second.

9

Preparation for the Mathematics examination brought to you by Kagiso Trust

ANSWERS EXERCISE 3. 1 QUESTION 1 1.1 -2 1.2 2 1.3 3 1.4 1

QUESTION 2 2.1 x = 7 2.2 x = 243 2.3 x = −2 2.4 x = 3 2.5 x = 4 2.6 x = 3 (reject x = −9 because negative numbers do not have logarithms) 2.7 x = 2 (similarly reject x = −4 ) 2.8 x = 2 (similarly reject x = −3 ) 2.9 x =

11 8

QUESTION 3 3.1 x = 2,8074 3.2 x = 0,5283 3.3 x = 0,3802 3.4 x = 0,1583 3.5 x = −1,2925 3.6 x = 1,2920 3.7 x = 56,1421 3.8 x = −0,0959 3.9 x = 0,0517

EXERCISE 3.2 1. n = 20years 6 months 2. n = 20 years 0 months 3. n = 33 years 5 months 4. t = 24 minutes 15 seconds 5. t = 12 minutes 39 seconds

KT Classroom Unit 3: Logarithms

Published on Sep 19, 2012

Kagiso Trust's KT Classroom: Unit 3_Logarithms