TOPIC: LOGARITHM – SOLVING PROBLEMS BASED ON LAWS OF LOGARITHM
CONTENT
- Logarithm of numbers (Index & Logarithmic Form)
- Laws of Logarithm
- Logarithmic Equation
- Change of Base
- Standard forms
- Logarithm of numbers greater than one
- Multiplication and divisions of numbers greater than one using logarithm
- Using logarithm to solve problems with roots and powers (no > 1)
- Logarithm of numbers less than one.
- Multiplication and division of numbers less than one using logarithm
- Roots and powers of numbers less than one using logarithm
Logarithm of numbers (Index & Logarithmic Form)
The logarithm to base a of a number P, is the index x to which a must be raised to be equal to P.
Thus if P = ax, then x is the logarithm to the base a of P. We write this as x = log a P. The relationship logaP = x and
ax =P are equivalent to each other.
ax =P is called the index form and logaP = x is called the logarithm form
Conversion from Index to Logarithmic Form
Write each of the following in index form in their logarithmic form
a) 26 = 64 b) 251/2 = 5 c) 44= 1/256
Solution
a) 26 = 64
Log2 64 = 6
b) 251/2 = 5
Log255=1/2
c) 4-4= 1/256
Log41/256 = -4
Conversion from Logarithmic to Index form
a) Log2128 = 7 b) log10 (0.01) = -2 c) Log5 2.25 = 2
Solution
a) Log2128 = 7
27 = 128
b) Log10 (0.01) = -2
10-2= 0.01
c) 5 2.25 = 2
1.52 = 2.25
Laws of Logarithms
a) let P = bx, then logbP = x
Q = by, then logbQ = y
PQ = bx X by = bx+y (laws of indices)
Logb PQ = x + y
:. Logb PQ = logbP + LogbQ
b) P÷Q = bx÷by = bx+y
LogbP/Q = x –y
:. LogbP/Q = logbP – logbQ
c) Pn= (bx)n = bxn
Logbpn = nbx
:. LogPn = logbP
d) b = b1
:. Logbb = 1
e) 1 = b0
Logb1 = 0
Example
Solve each of the following:
a) Log327 + 2log39 – log354
b) Log35 – log310.5
c) Log28 + log23
d) Given that log102 = 0.3010 log103 = 0.4771 and log105 = 0.699 find the log1064 + log1027
Solution
a) Log327 + 2 log39 – log354
= log3 27 + log3 92 –log354
= log3 (27 x 92/54)
= log3 (271 x 81/54) = log3 (81/2)
= log3 34/ log32
= 4log3 3 – log3 2
= 4 x (1) – log3 2 = 4 – log3 2
= 4 – log3 2
b) log3 5 – log3 10.5
= log3 (13.5) – Log310.5 = log3 (135/105)
= log3 (27/21) = log3 27 – log3 21
= log3 33 – log3 (3 x 7)
= 3log3 3 – log3 3 -log37
= 2 – Log3 7
c) Log28 + Log33
= log223+ log33
= 3log22 + log33
= 3 +1 = 4
d) log10 64 + log10 27
log10 26 + log1033
6 log10 2 + 3 log10 3
6 (0.3010) + 3(0.4771)
1.806 + 1.4314 = 3.2373.
EVALUATION
- Change the following index form into logarithmic form.
(a) 63= 216 (b) 33 = 1/27 (c) 92 = 81
2. Change the following logarithm form into index form.
(a) Log88 = 1 (b) log ½¼ = 2
3. Simplify the following
a) Log55 + log52 b) ½ log48 + log432 – log42 c) log381
4. Given that log 2 = 0.3010, log3 0.4770, log5 = 0.6990, find the value of log 6.25 + log1.44
Logarithmic Equation
Solve the following equation:
- a) Log10 (x2 – 4x + 7) = 2
- b) Log8 (r2 – 8r + 18) = 1/3
Solution
- a) Log10 (x2 – 4x + 7) = 2
x2 – 4x + 7 = 102 (index form)
x2 – 4x + 7 = 100
x2 – 4x + 7 – 100 = 0
x2 – 4x – 93 = 0
Using quadratic formula
x = – b ±√b2– 4ac
2a
a = 1, b = – 4, c = – 93
x = – (- 4) ± √(- 4) 2 – 4 x 1 x (- 93)
2 x 1
= + 4 ± √16 + 372
2
= + 4 ± √388/2
= x = 4 +√ 388/2 or 4 – √388/2
x = 11.84 or x = – 7.85
2) Log8 (x2 – 8x + 18) =1/3
x2 – 8x + 18 = 81/3
x2 – 8x + 18 = (2)3X1/3
x2 – 8x + 18 =2
x2 – 8x 18 – 2 = 0
x2 – 8x + 16 = 0
x2 – 4x – 4x + 16 = 0
x(x – 4) -4 (x – 4) = 0
(x – 4) (x – 4) = 0
(x – 4) twice
x = + 4 twice
Change of Base
Let logbP = x and this means P = bx
LogcP = logcbx = x logcb
If x logcb = logcP
x = logcP
logc b
:. logcP = logcP
logcb
Example :
Shows that logab x logba = 1
Logab = logcb
logca
Logba = logca
logcb
:. logab x logba = logcb x logca
logca + logcb = 1
Evaluation
Solve (i) Log3 (x2 + 7x + 21) = 2 (ii) Log10 (x2 – 3x + 12) = 1
(iii) 52x+1 – 26(5x) + 5 = 0 find the value of x
Logarithm of numbers greater than one
Numbers such as 1000 can be converted to its power of ten in the form 10n where n can be term as the number of times the decimal point is shifted to the front of the first significant figure i.e. 10000 = 104
Number Power of 10
- 102
- 101
- 100
- 10-3
- 10-1
Note: One tenth; one hundredth, etc are expressed as negative powers of 10 because the decimal point is shifted to the right while that of whole numbers are shifted to the left to be after the first significant figure.
A number in the form A x 10n, where A is a number between 1 and 10 i.e. 1 < A < 10 and n is an integer is said to be in standard form e.g. 3.835 x 103 and 8.2 x 10-5 are numbers in standard form.
Examples
Express the following in standard form
1) 7853 2) 382 3) 0.387 4) 0.00104
Solution
1) 7853 = 7.853 x 103
2) 382 = 3.82 x 102
3) 0.387 = 3.87 x 10-1
4) 0.00104 = 1.04 x 10-3
Base ten logarithm of a number is the power to which 10 is raised to give that number e.g.
628000 = 6.28 x105
628000 = 100.7980 x 105
= 100.7980 + 5
= 105.7980
Log 628000 = 5.7980
Integer Fraction (mantissa)
If a number is in its standard form, its power is its integer i.e. the integer of its logarithm e.g. log 7853 has integer 3 because 7853 = 7.853 x 103
Examples: Use tables (log) to find the complete logarithm of the following numbers.
(a) 80030 (b) 8 (c) 135.80
Solution:
(a) 80030 = 4.9033
(b) 8 = 0.9031
(c) 13580 = 2.1329
Multiplication and Division of numbers greater than one using logarithm
To multiply and divide numbers using logarithms, first express the number as logarithm and then apply the addition and subtraction laws of indices to the logarithms. Add the logarithm when multiplying and subtract when dividing.
Examples: Evaluate using logarithm.
- 4627 x 29.3
- 8198 ÷ 3.905
- 48.63 x 8.53
15.39
Solutions
- 4627 x 29.3
No Log
4627 3.6653
29.3 + 1.4669
Antilog → 135600 5.1322
∴ 4627 x 29.3 = 135600
To find the Antilog of the log 5.1322 use the antilogarithm table:
Check 13 under 2 diff 2 (add the value of the difference) the number is 0.1356. To place the decimal point at the appropriate place, add one to the integer of the log i.e. 5 + 1 = 6 then shift the decimal point of the antilog figure to the right (positive) in 6 places.
= 135600
- 819.8 x 3.905
No Log
819.8 2.9137
3.905 0.5916
antilog →
209.9 2.3221
∴ 819.8 ÷ 3.905 = 209.9
- 48.63 x 8.53
15.39
No Log
48.63 1.6869
8.53 +0.9309
2.6178
÷ 15.39 -1.1872
antilog → 26.95 1.4306
∴ 48.63 ÷ 8.53 = 26.96
15.39
Evaluation:
- Use table to find the complete logarithm of the following:
(a) 183 (b) 89500 (c) 10.1300 (d) 7
2 Use logarithm to calculate. 3612 x 750.9
113.2 x 9.98
Using logarithms to solve problems with powers and root (numbers greater than one).
Examples:
Evaluate
(a) 3.533 (b) 4 40000 (c) 94100 x 38.2
5.683 x 8.14 correct to 2s.f.
Solution
No. Log_____
3.533 0.5478 x 3
44.00 1.6434
∴ 3.533 = 44.00
(b) 4 4000
No. Log_____
4 4000 3.6021 ÷ 4
7.952 0.9005
∴ 4 4000 = 7.952
(c) 94100 x 38.2
5.6833 x 8.14
Find the single logarithm representing the numerator and the single logarithm representing the denominator, subtract the logarithm then find the anti log.
(Numerator – Denominator).
No Log
94100 4.9736 ÷ 2 = 2.4868
38.2 1.5821
Numerator 4.0689 → 4.0689
5.683 0.7543 x 3 = 2.2629
8.14 0.9106
Denominator 3.1735 → 3.1735
7.859 0.8954
∴ 94100 x 38.2 = 7.859
5.683 x 8.14
~ 7.9 (2.sf)
Evaluation:
Evaluate using logarithm.
95.3 x 318.4
1.295 x 2.03
Logarithm of number less than one
To find the logarithm of number less than one, use negative power of 10 e.g.
0.037 = 3.7 x 10-2
10 0.5682 x 10-2
10 0.5682 + (-2)
10-2 5682
Log 0.037 = 2 . 5682
2 . 5682
Integer decimal fraction (mantissa)
Example:
Find the complete log of the following.
(a) 0.004863 (b) 0.853 (c) 0.293
Solution
Log 0.004863 = 3.6369
Log 0.0853 = 2.9309
Log 0.293 = 1.4669
Evaluation
- Find the logarithm of the following:
(a) 0.064 (b) 0.002 (c) 0.802
Using logarithm to evaluate problems of Multiplication, Division, Powers and roots with numbers less than One
Examples:
- 0.6735 x 0.928 2. 0.005692 ÷ 0.0943 3. 0.61043
4 0.00083 5. 3 0.06642
Solution
- 0.6735 x 0.928
No. Log.___
0.6735 1.8283
0.928 1.9675
0.6248 1.7958
∴ 0.6735 x 0.928 = 0.6248
b. 0.005692 ÷ 0.0943
No Log
0.005692 3.7553
÷ 0.0943 2.9745
0.06037 2.7808
c. 0.61043
No Log_____
0.61043 1.7856 x 3
0.2274 1.3568
∴ 0.61043 = 0.2274
∴ 0.005692 ÷ 0.943 = 0.6037
4 0.00083
No. Log._____
4 0.00083 4.9191 ÷ 4
0.1697 1.2298
∴ 4 0.06642 = 0.1697
- 3 0.6642
No. Log.____________
3 0.6642 2.8223 ÷ 3
2.1 + 1 + 0.8223 ÷ 3
3 + 1 .8223 ÷ 3
1 + 0.6074
0.405 1.6074
3 0.6642 = 0.405
Note: 3 cannot divide 2 therefore subtract 1 from the negative integer and add 1 to the positive decimal fraction so as to have 3 which is divisible by 3 without remainder.
Evaluation:
Evaluate using logarithm tables:
(1) √12.3 x 0.00343
132.5
(2) 23.97 x 0.7124
- x 52.18
General Evaluation
- Solve the logarithmic equation: Log4 (x2 + 6x + 11) = ½
- Log2 (x2– 2) =log2(x-1) + 1
- Evaluate 5 (0.1684)3
6.28 x 304
981
163/2 x 82/3
321/5
Weekend Assignment
1.) If log81/64 = x, find the value of x (a) 2 (b) 1 (c) -3 (d) -4.
2.) Solve 9(1 – x) = (1/27) x+1 (a) -5 (b) -1 (c) 1 (d) ½
Use a table to find the log of the following:
3.) 900 (a) 3.9542 (b) 1.9542 (c) 2.9542 (d) 0.9542
4.) 0.000197 (a) 4.2945 (b) 4.2945 (c) 3.2945 (d) 3.2945
5.) Use the antilog table to write down the number whose logarithm is 3.8226.
(a) 0.6646 (b) 0.06646 (c) 0.006646 (d) 66.46
Theory
- Find the value of x for which log10 (4x2 + 1) -2 log10 x – log10 2 = 1 is valid.
(2.) Evaluate using logarithms.