Week 2 – SS3 First Term Mathematics Notes

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)    3^x = 1/81        d) 10 ^x = 1/0.001
e)    4/2^x = 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)    3^x = 1/81
    3^x = 1/3⁴
    3^x = 3 ^-4
    X = -4
d)    10^x = 1/0.001
    10 ^x = 1000
    10 ^x = 10 ³
    10^x = 10 ³
X = 3
e)    4/2^x = 64 ^x
    4÷2^x = 64 ^x
    2² ÷2^x = 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 5^x = 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 3^x = 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

1. Solve for X : (0.25) ^{X + 1} = 16 (a) -3 (b) 3 (d) 4 (d) -4
2. Solve for X : 3(3)^X = 27 (a) 3 (b) 4 (c) 2 (d) 5
3. Solve the exponential equation : 2^2X + 2^X+1 – 8 = 0 (a) 1 (b) 2 (c) 3 (d) 4
4. The second value of X in question 3 is (a) -1 (b) 1 (c) 2 (d) It has no solution
5. Solve for X : 10^-X = 0.000001 (a) 4 (b) 6 (c) -6 (d) 5

Theory
Solve the following exponential equations
(1.) (3^X)² + 2(3^X )– 3 = 0 (2). 5^2X+1– 26(5^X) + 5 = 0