Laws of Indices
Further Mathematics SS 1 FIRST TERM
WEEK ONE
LAWS OF INDICES:
CONTENT:
- Laws of Indices
- Application of Indices Linear Equations
- Application of Quadratic Equations
LAWS OF INDICES
There are laws governing the use of indices. These are useful in other subjects. They are;
- an x am = an+m
Therefore
a3 x a3 = a x a x a x a x a x a = a6 i.e a3+3 = a6
In general, when multiplying indices with same base, you add the power
- an ÷ am = an-m .
a5 ÷ a4 = a5-4 = a1
Also, a5 ÷ a4 = = a
In general, when dividing indices with the same base, you subtract the power
- (an)m= an×m = amn
(a3)2 = a3×2 = a6
- a0 =1
a5 ÷ a5 = a5-5 = a0
Also, a5 ÷ a5 = = 1
This implies that anything to power zero is equal to 1, i. e 50 = 1, 20 = 1
- a-n= 1/an
Consider a5 ÷ a6 = a5-6 = a-1
But a5÷ a6 = =
a-1 =
In the same way a-2=,
- a n/m =n = )
Consider (8) 1/3 = (23) 1/3 = (23) 1/3 = 2
= 2,
Example1:
(i) (a5 x a6) /a5 (ii) (26 ÷27 x 24)1/3 (iii) 321/5 (iv) 216 ÷ 34
Solution:
- (a5 x a6) /a5 = a3 +6 -5 = a9-5 = a4
- (26÷ 27 x 24)1/3 = (26-7+4)1/3
Re arrange the indices
(26+4-7)1/3 = (23)1/3 = 2
iii. 321/5 = 321/5 but 32 = 25
(25) 1/5 = 25x 1/5 = 2
iv. 216 ÷ 34
216 = 23 x 33
23 x 33 ÷ 34 = 23 x 33-4 = 23 x 3-1
23/3 0r 8/3
Example2:
Simplify
Solution
= = =
CLASS ACTIVITIES:
- Evaluate each of the following
(a) 80 (b) 5-1 (c) 82/3 (d) (x3)-2/3 (e) (43)5
- Evaluate each of the following
(f) (625)-1/4 (g) 642/3 (h) 91/3 x 91/6 (i) 36 ÷ 37 x 22 (j) (1000)-5/3
SUB TOPIC: APPLICATION OF THE LAWS OF INDICES.
Examples:
- Solve (1/2)x = 8
Solution:
(1/2)x = (2-1)x = 2-x
8 = 23 since we have same base, then –x = 3.
Multiply the equation by (-1)
X = -3
- Solve the equation 8x = 0.25
Solution:
8x = (23)x = 23x
0.25 = 25/100 = ¼ = (1/2)2 = (2-1)2 = 2-2
23x = 2-2
3x = -2
X = -2/3
- Solve for x in the equation: (0.25)x+1 = 16
(1/4)x+1 24
(2-2)x+1 = 24
2-2x-2 = 24
Equate the power
-2x – 2 = 4
-2x = 4 + 2 = 6
X = -6/2 = -3
X = -3
If 10-x = 0.001. what is the value of x?
0.001 = 10-3
–x = -3
x = 3
- If 25(5x) = 625, what is x.?
(52)5x = 54
510x = 54
10x = 4
X = 4/10 0r 2/5
CLASS ACTIVITIES:
Solve for x in the following equations
- 3x = 81
- 2x = 32
- 9x = 1/729
- 25(5x) = 625
- 2x x 4-x = 2
SUB TOPIC: APPLICATION OF INDICES LEADING TO QUADRATIC EQUATION
Some exponential equations will lead to quadratic equations as you will see in the following examples.
- 52x -30 x 5x + 125 =0
Solution:
Re-write the equation
(5x)2 -30 x 5x + 125 =0
Let 5x = p, then
P2 – 30p + 125 = 0
Solve for p by factorization
(p-5)(p-25) = 0
P – 5 = 0 or p – 25 = 0
Then p = 5 or 25,
Recall that p =5x
Therefore 5x = 51, then x = 1
or 5x = 25 this means that 5x = 52, x = 2
Solve the equation 22x + 4(2x) – 32 = 0
22x + 4(2x) – 32 = 0
(22x)2 + 4(2x) – 32 = 0
Let 2x = y, then
y2 + 4y – 32 = 0
(y + 8)(y – 4) = 0
y = 4 or -8
Then 2x = 22 or 2x = -8. But this (2x = -8) has no solution
Therefore x = 2
- Solve for x in the equation
32(x-1) – 8(3(x-2)) = 1
Solution:
Re write the equation
32 x 3-2 – 8 x 3x x 3-2 – 1 = 0
32x x 1/32 – 8 x 3x x 1/32 – 1 = 0
Multiply the equation by 32
32x – 8(3x) – 32 = 0
Lep p = 3x
P2 – 8p – 9 = 0
(p-9)(p+1) = 0
P = 9 or -1
Recall that p = 3x
3x = 32, 3x = -1 has no solution
x = 2
CLASS ACTIVITIES:
Solve the following equations
- 22x – 5(2x) + 4 = 0
- 32x+1 + 26(3x) – 9 = 0
- 22x – 6(2x) = -8
- 72x – 2 x (7x) = -1
- 2x+3 – 15 = 211-x
PRACTICE EXERCISE:
Objective Test:
Choose the correct answer from the options
- Simplify (105)0 (a) 0 (b) 1 (c) 5 (d) 3 (e)-1
- Evaluate 3432/3 (a) 7 (b) 49 (c) 343 (d) 3 (e) 9
- Simplify (28 x 4-3) / 26 (a) 1/16 (b) 16 (c) 220 (d) ¼ (e) 2/4
- Solve the equation 3-x = 243, x = ? (a) 5 (b) 3 (c) 4 (d) -5 (e) -3
- Solve the equation 32x – 9 = 0, x = ? (a) ±3 (b) 2 (c) -2 (d) 1 (e) 0
Essay questions:
- Simplify (a) 4
- (1/3)4 x 36 ÷ (2/3)2
- Solve for x, if 125x-1 = 252x-3
- If 92x+1 = 81x-2/3x, what is x?
- Find the value of x satisfying
- 32x – 30(3x) + 81 = 0
WEEKEND ASSIGNMENT:
Simplify the following
- If =
- Solve
- If
KEY WORDS:
- INDEX (plural INDICES)
- INDEX FORM
- BASE