Theory of Logarithms: Laws of Logarithms and application of Logarithmic equations and indices

SUBJECT: MATHEMATICS

CLASS: SS 3

TERM: FIRST TERM

WEEK 1                                                                                                                      DATE:_______________________

THEORY OF LOGARITHMS AND LAWS OF LOGARITHMS

In general, the logarithm of a number is the power to which the base must be raised in order to give that number. i.e if y=n^x, then x = log_ny. Thus, logarithms of a number to base ten is the power to which 10 is raised in order to give that number i.e if y =10^x, then x =log₁₀y. With this definition log₁₀100= 3 since 10³= 1000 and log₁₀100 = 2 since 10²=100.

Examples:

1.   Express the following in logarithmic form

a)   2^-6 = 1/64     b) 3⁵ =243    c) 5³ = 125    d) 10⁴ = 10,000

Solutions

1.   (a) 2^-6 = 1

64

∴log₂ (1/64) = -6

(b)3⁵ = 243

∴ log₃243 =5

(c)5³ =125

∴ log₅125 = 3

(d) 10⁴ = 10,000

∴log₁₀10000 = 4

2.   Express the following in index form

a)   Log₂(1/8) = -3     (b) Log₁₀(1/1000) = -2     (c) Log₄64      (d) Log₅625

(e) Log₁₀1000

Solutions

a)   Log₂ (1/8)= -3

Then 2^-3 = 1/8

b)   Log₁₀(1/100) = -2

Then 10^-2 = 1/100

c)    Let Log₄64 = k

Then 4^k  = 64

Simplify 64; 4^k = 4³

Then k = 3

d)   Let Log₅625 = m

Then 5^m = 625

5^m = 5⁴

m = 4

e)   Let Log₁₀1000 = p

Then 10^p = 1000

10^p = 10³

P = 3

Evaluation: Evaluate the following logarithms

1. Log₄8   2. Log₆216   3. Log₈0.0625   4. Log3125 5. Log5625 6. Log0.3125

Solutions

1) Log48 = log21 * 2

2) Log6216 = log26216

3) Log80.0625 =log 8 + 0.0625

4) Log3125 = log314

5) Log5625 = log53125

Basic Laws of Logarithms

1. Log mn = Log m + Log n
2. Log (m/n) = Log m – Log n
3. Log m^p  = pLog m
4. Log 1 = 0
5. Log_mm = 1
6. Log (1/m)ⁿ= Log m^-n = -nLog m
7. Log m”n = n Log m
8. Log (m/n) = (-n)/(mn) Log m + 1/m Log n
9. 1. Let logm x=a and logxn=b, then logmx = a+b

Change of base

Log _mn = Log_an

Log_am

Examples:

1.   Express as logarithm of a single number 2Log3 + Log6

solution:

2log3 + log 6

= log3²+ log 6 = log 9 + log 6

= Log 9 x6 = Log 54

2.    Simplify Log 8 ÷ Log 4

Solution:

Log 8 ÷ log 4 = log 2³ = 3log2 = 3/2

log 2²    2log2

3.   Evaluate 3log2 + log 20 – log 1.6

Solution

= log2³ + log 20 – log(16/10)

= log 8 + log 20 – log (8/5)

= log (8 x 20÷ 8/5)

= log (8 x 20 x 5 ÷ 8)

= log 100 = log10^{2 =} 2log 10 = 2

Evaluation: 

1.   Simplify the following: a. ½ log 25  b. 1 + log 3  c. log 8 + log 4

2.   Evaluate log 45 – log 9 + log 20

3.   Given that log 2 = 0.3010, log 3 = 0.4771 and log 7 = 0.8451, evaluate a. log 42  b. log 35

4.   Solve 2^2x  – 10 x 2^x+ 16 = 0

Solution:

22x – 10×2 + 16 = 0

Simplify the equation by removing all powers of 2 and x. Then

Further Application of Logarithms using tables:

Examples:

Use the tables to find the log of:

(a)  37    (b) 3900 to base ten

Solutions

1.   37 = 3.7 x 10

=3.7 x 10¹(standard form)

=10^{0.5682 + 1} x 10¹ (from table)

=10^1.5682

Hence log₁₀37 = 1.5682

2.   3900 = 3.9 x 1000

=3.9 X 10³ (standard form)

=10^0.5911 x 10³ (from table)

=10^{0.5911 + 3}

=10^3.5911

Therefore log₁₀3900 = 3.5911

Evaluate the following using tables

1.   4627 x 29.3

2.   819.8 ÷ 3.905

3.

48.63 x8.53

15.39 x 3.52

Solutions

1.   4627 x 29.3

No         Log

4627  3.6653

29.3  1.4669

135600     5.1322

4627 x 29.3 = 135600 (4 s.f)

2.   819.8 ÷ 3.905

No        Log

819.8  2.9137

3.905   0.5916

209.9    2.3221

Therefore 819.8 ÷ 3.905 = 209.9

3.      48.63 x 8.53

15.39 x 3.52

No       log

48.63 1.6869

8.53    0.9309

Numerator    2.61782.6178

15.39          1.1872

3.52    0.5465

Denominator1.7337          1.7337

7.658                    0.8841

Therefore 48.63 x 8.53 = 7.658

15.39 x 3.52

Evaluation

Use logarithms tables to calculate

1.    36.12 x 750.9          2. 319.63 x 12.28² x 74

113.2 x 9.98

General evaluation

1.   Change the following to logarithms form

a.    25 ^½ = 5     b. (0.01)² = 0.0001

2.   Change the following to index form

a.    Log31 = 0   b. Log₁₅225 =2

3.   Evaluate the following using logarithms tables

a.  69.7 x 44.63

25.67

b.   17.9 x 3.576 x 98.14

c.

(35.61)² x 5.62

                  ³√143.5

Reading assignment: NGM for SSS BK1 pg 18 – 22 and ex. 1c Nos 19 – 20 page 22

Weekend Assignment

1.   Find the log of 802 to base 10 (use log tables) (a) 2.9042 (b) 3.9040 (c) 8.020 (d)1.9042

2.   Find the number whose logarithm is 2.8321 (a) 6719.2 (b) 679.4 (c) 0.4620 (d) 67.92

3.

What is the integer of the log of 0.000352 (a) 4 (b) 3 (c) 4 (d)3

4.   Given that log₂(1/64) = m, what is m ? (a) -5 (b) -4 (c) -6 (d) 3

5.   Express the log in index form: log₁₀10000 =4 (a) 10³ = 10000 (b) 10^-4 = 10000 (c) 10⁴ = 10000

(d) 10⁵ =100000

6. Simplify

log10 (2/25) – log 10( 4/125)

a. log 10 2/25 = 0.3010½b. log 10

4/125=½c. log 10

16/(5 x 25)=1/5

7. Evaluate: log 10 (67) + 1/log10 (81) – 2

a. log 10 67 = 3½b. log 10 81 = 4 c. 1/log 10 81 – 2= 1/4-2= -3/4 d. log 10 67+ 1/log 10 81 – 2 = 3+ 1/-3 –2=-1.5

8. Change the following to index form: log2 (7) + log 0.0081 (a) 5 (b) 4 (c) 6 (d) 7

a. log 2 7=1½b. log 0.0081=∴c.ln2x + ln1/8100=5d.ln(7) + ln0.0081 = 1+ ∴-ln100 = 6

9. Evaluate: 12log 10 (1000 x 473) – 15log 10 (0.8)​

a. 12log 10 (1000 x 473)=12½(x2/3) -15 log10(0.8)=12½(-1/3) -15 ∴=1-5 = 10

b. 12log10(1000×473) – 15log10(0.8)= 12½(-1/3x1000x4/3) – 15∴(-1/8)=10-15 ∴= 5

c. 12log 10 (1000 x 473)-15 log 10 (0.8) = 100-120 = -20

10. Given that log2(64) = 4, find the value of 64÷log 2 (a) 1 (b) 2 (c) 4 (d) 16

Theory

1.

Evaluate using logarithm table 6.28 x 304

981

and express your answer in the form A X 10ⁿ, where A is a number between1 and 10 and n is an integer.

2.   Use logarithm table to calculate 6354 x 6.243 correct to 3 s.f.

39253.3

3. Use logarithm table to calculate 826 ÷ 4.115 correct to 3 sf

197.62

4. Simplify the following expression: ln 5 + ln 2 – ln 6 + ln8 – 1/ln10 (a

16.76 x 0.0323

log10 16.76 = 2 x 0.0323 =0.0988 (2 sf)

ln(16.76) + ln2 – ln6 + ln8 – 1/ln10=4+1/-3-4+1/+10= 4-7/31= 16.