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 ( USE THE DISCUSSION BOX AT THE BOTTOM TO SUBMIT YOUR ANSWER FOR DISCUSSION AND APPRAISAL)
Solve the following exponential equations.
- a) 2 2x+ 1 – 5 (2 x) + 2 = 0
- b) 3 2x – 4 (3 x+1) + 27 = 0
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