# INDICES AND LOGARITHMS - Mathematics Form 2 Notes

## Indices

### Index and Base Form

• The power to which a number is raised is called index or indices in plural.
25=2 × 2 × 2 × 2 × 2
• 5 is called the power or index while 2 two is the base.
100 =
102
• 2 is called the index and 1 0 is the base.

### Laws of Indices

• For the laws to hold the base must be the same.

Rule 1
• Any number, except zero whose index is 0 is always equal to 1
Example
50 = 1
100000000000000000=1

Rule 2
• To multiply an expression with the same base, copy the base and add the indices.
am × an = am+n
Example
52 × 53=55
= 3125

Rule 3
• To divide an expression with the same base, copy the base and subtract the powers.
am ÷ an = am−n
Example
95 ÷ 92 = 93

Rule 4
• To raise an expression to the nth index, copy the base and multiply the indices
amxn = amn
Example
(53)2
=53×2=56

Rule 5
When dealing with a negative power, you simply change the power to positive by changing it into a fraction with 1 as the numerator.
a−m  1
am
Example
22=
22
=1/4
Example

Evaluate:
1. 23×23 = 2(3+−3)
=20
=1
2. (2/3)−2=(1)2
(2/3)2
1
4/9
=1 ÷ 4/9
=1 × 9/421/or (2/3)−2 = (3/2)= 9/4

### Fractional Indices

• Fractional indices are written in fraction form. In summary if an = b. a is called the nth root of b written as nb.

Example

271/3 =∛27 = 3

163/4 = (4√16)3 = 23= 8

4−½  1
4½

=
√4
= ½

## Logarithms

• Logarithm is the power to which a fixed number (the base) must be raised to produce a given number. In summary the expression am = n is written as loga n = m.
am = n is the index notation while loga n = m is the logarithm notation.

Examples

 Index notation Logarithm form 22 =4 log24 = 2 9½ = 3 log93 = ½ bn = m logbm =n
• Reading logarithms from the tables is the same as reading squares square roots and reciprocals.
• We can read logarithms of numbers between 1 and 10 directly from the table. For numbers greater than 10 we proceed as follows:
• Express the number in standard form, A × 10n.Then n will be the whole number part of the logarithms.
• Read the logarithm of A from the tables, which gives the decimal part of the logarithm. Then add it to n which is the power of 10 to form the positive part of the logarithm.

Example

Find the logarithm of: 379

Solution

In standard form
= 3.79 x
102
Log 3.79 = 0.5786
Therefore the logarithm of 379 is 2 + 0.5786= 2.5786

Note: The whole number part of the logarithm is called the characteristic and the decimal part is the mantissa.

### Logarithms of Positive Numbers Less Than 1

Example

Log to base 10 of 0.034

Solution

We proceed as follows:

• Express 0.034 in standard form, i.e., A X10n.
• Thus 0.034 = 3.4 x 10-2
• Log 3.4 from the tables is 0.531 5
• Hence 3.4 x 10-2 = 100.5315 × 10-2
Using laws of indices add 0.531 5 + -2 which is written as ̅2.5315.
• It reads bar two point five three one five. The negative sign is written directly above two to show that it’s only the characteristic is negative.

Example

Find the logarithm of: 0.00063

Solution

= 6.3 ×10-3 (Find the logarithm of 6.3)
= 100.7993 ×10-3
= -3 + 0.7993
= ̅3
.7993

## Antilogarithms

• Finding antilogarithm is the reverse of finding the logarithms of a number.
• For example the logarithm of 1000 to base 10 is 3. So the antilogarithm of 3 is 1000.
• In algebraic notation, if

Log x = y then antilog of y = x.

Example

Find the antilogarithm of ̅2 .3031

Solution

Let the number be x
x = 10̅2 .3031
= 10-2+0.3031
= 10-2 x 100.3031(Find the antilog, press shift and log then key in the number)
= 10-2 x 2.01
= 1/100 x 2.01
=2.01/100
= 0.0201

Example

Use logarithm tables to evaluate:
456 x 398
271

Solution

 Number Standard form logarithm 456 398 271669.7 4.56 x 102 3.98 x 1022.71 x 1026.697 x102 ← antilog 2.6590 +2.5999   5.2589 −2.5330 2.8259

= 669.7

To find the exact number find the antilog of 2.8259 by letting the characteristic part to be the power of ten then finding the antilog of 0.8259

Example
Operations involving bar

Evaluate 415.2 x 0.0761
135
Solution

 Number logarithm 415.20.07611352.341 x 10-1 ←antilog 2.6182̅2.8814 +1.49962.1303−̅1.3693 0.2341

= 0.2341

Example

√0.945 = (9.45 x 10-1)½
= (10̅1.9754 x ½ )

Note;
In order to divide ̅1.9754 by 2, we write the logarithm in search away that the characteristic is exactly divisible by 2.

If we are looking for the nth root, we arrange the characteristic to be exactly divisible by n
̅1.9754 = −1 + 0.9754
= −2 + 1.9754
Therefore,
1/2(̅1.9754) = (−2 + 1.9754)
= −1 + 0.9877
= ̅1.9877
Find the antilog of ̅1.9877 by writing the mantissa as power of 1 0 and then find the antilog of characteristic.
= 9.720 X 10-1
= 0.9720

Example

30.0618

Solution

 Number Logarithm 3√0.0618 = 0.06181/3   3.954 ×10-1     ←antilog ̅2.7910 × 1/3( ̅3 + 1.7910) × 1/3= ̅1.5970

= 0.3954

## Past KCSE Questions on the Cubes, Cubes Roots, Reciprocals, Indices and Logarithms.

1. Use logarithms to evaluate

2. Find the value of x which satisfies the equation.
16x2 = 84x−3
3. Use logarithms to evaluate

4. Use logarithms to evaluate
55.9 ÷ (0.2621 x 0.01 177)1/5
5. Simplify 2x x 52x÷2x
6. Use logarithms to evaluate
(3.256 x 0.0536)1/3
7. Solve for x in the equation
32(x -3) ÷8(x-4) = 64 ÷2x
8. Solve for x in the equation
81
2x x 27x  = 729
9x
9. Use reciprocal and square tables to evaluate to 4 significant figures, the expression:
1  + 4.3462
24.56
10. Use logarithm tables, to evaluate

11.  Find the value of x in the following equation
49(x +1) + 7(2x) = 350
12. Use logarithms to evaluate
13. Find the value of m in the following equation
(1/27m x (81)-1 = 243
14. Given that P = 3y express the equation 3(2y-1) + 2 x 3(y-1) = 1 in terms of P hence or otherwise find the value of y in the equation 3 (2y – 1) + 2 x 3(y-1) = 1
15. Use logarithms to evaluate
16. Use logarithms to evaluate

17. Solve for x in the equation
x =   0.0056½
1.38 x 27.42

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