WEEK 2
TOPIC: INDICIAL/EXPONENTIAL EQUATION
CONTENT:
- Exponential Equation of Linear Form
- Exponential Equation of Quadratic Form
Under exponential equation, if the base numbers of any equation are equal, then the power will be equal&vice versa.
Exponential Equation of Linear Form
Solve the following exponential equations
a) (1/2) x = 8 b) (0.25) x+1 = 16
c) 3x = 1/81 d) 10 x = 1/0.001
e) 4/2x = 64 x
Solution
a) (1/2) X = 8
(2 -1) x = 2 3
2 –x = 2 3
-x = 3
X = – 3
b) (0.25) x+1 = 16
(25/100) x+1 = 16
(1/4) x+1 = 4 2
(4 -1) x+1 = 4 2
– x – 1 = 2
– x = 2 + 1
– x = 3
X = – 3
c) 3x = 1/81
3x = 1/34
3x = 3 -4
X = -4
d) 10x = 1/0.001
10 x = 1000
10 x = 10 3
10x = 10 3
X = 3
e) 4/2x = 64 x
4÷2x = 64 x
22 ÷2x = 64 x
2 2-x = (2 6) x
2 2-x = 2 6x
2-x = 6x
2=6x+x
2 = 7x
Divide both sides by 7
2/7 = 7x/7
X = 2/7
EVALUATION
Solve the following exponential equations
a) 2 x = 0.125 b) 25 (5x) = 625 c) 10 x = 1/100000
Exponential Equation of Quadratic Form
Some exponential equation can be reduced to quadratic form as can be seen below.
Example : Solve the following equations.
a) 2 2x – 6 (2 x) + 8 = 0
b) 5 2x + 4 X 5 x+1 – 125 = 0
c) 3 2x – 9 = 0
Solution
a) 2 2x – 6 (2 x) + 8 = 0
(2 x)2 – 6 (2 x) + 8 = 0
Let 2 x = y.
Then y2 – 6y + 8 = 0
Then factorize
Y 2 – 4 y – 2y + 8 = 0
Y (y – 4) -2 (y -4) = 0
(y -2) (y – 4) = 0
Y – 2 = 0 or y – 4 = 0
Y = 2 or y= 4
Y = 2, 4
Since 2 x = y, and y = 2
2 x = 2
2 x = 2 1
x = 1
Since 2 x = y and y = 4
2 x = 4
2 x = 2 2
N = 2
X = 1 and 2
b) 5 2x + 4 X 5 x+1 – 125 = 0
(5 x) 2 + 4 X (5 x X 5 1) – 125 = 0
Let 5 x = p
P 2 + 4 X (p X 5) – 125 = 0
P2 + 4 (5p) – 125 = 0
P2 + 20p – 125 = 0
Then Factorise p2 + 25p – 5p – 125 = 0
P (p + 25) -5 (p + 25) = 0
(p – 5) (p + 25) = 0
P – 5 = 0 p + 25 = 0
P = 5 or p = – 25
Since 5x = p, p = 5
5 x = 5 1
X = 1
5x = -25 (Not simplified)
1) 3 2x – 9 = 0
(3 x) 2 – 9 = 0
Let 3 x = a
a 2 – 9 = 0
a 2 = 9
a = ±√9
a = ± 3
a = 3 or – 3
Since 3 x = a, when a = 3
3 x = 3 1
X = 1
Since 3x = a, when a = -3
3 x = – 3 (Not a solution)
EVALUATION
Solve the following exponential equations.
a) 2 2x+ 1 – 5 (2 x) + 2 = 0
b) 3 2x – 4 (3 x+1) + 27 = 0
Reading Assignment : Further Maths Project Book 1(New third edition).Chapter 2 pg. 6- 10
WEEKEND ASSIGNMENT
- Solve for X : (0.25) X + 1 = 16 (a) -3 (b) 3 (d) 4 (d) -4
- Solve for X : 3(3)X = 27 (a) 3 (b) 4 (c) 2 (d) 5
- Solve the exponential equation : 22X + 2X+1 – 8 = 0 (a) 1 (b) 2 (c) 3 (d) 4
- The second value of X in question 3 is (a) -1 (b) 1 (c) 2 (d) It has no solution
- Solve for X : 10-X = 0.000001 (a) 4 (b) 6 (c) -6 (d) 5
Theory
Solve the following exponential equations
(1.) (3X)2 + 2(3X )– 3 = 0 (2). 52X+1– 26(5X) + 5 = 0