Week 2 – SS1 First Term Further Mathematics Notes

 WEEK TWO
TOPIC: INDICIAL AND EXPONENTIAL EQUATIONS
CONTENT

• Exponential Equation of Linear Form
  
• Exponential Equation of Quadratic Form
  

Exponential Equation of Linear Form
Under exponential equation, if the base numbers of any equation are equal, then the power will be equal & vice versa.

 Examples
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 b)    (0.25) ^x+1 = 16
(2 ^-1) ^x = 2 ³ (25/100) ^x+1 = 4²
2 ^–x = 2 ³ (1/4) ^x+1 = 4²
-x = 3 (4^-1) ^{x + 1} = 4²
x = – 3 4 ^{– x – 1} = 4²
– x – 1 = 2
     – x = 2 + 1
– x = 3
X = – 3

 c)    3^x = 1/81 d) 10 ^x = 1/0.001
3^x = 1/3⁴ 10 ^x = 1000
    3^x = 3 ^-4 10 ^x = 10 ³
     x = -4 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 When y = 4 then, and     When y = 2 then,
(2^x)² – 6 (2^x) + 8 = 0 2 ^x = 4 2 ^x = 2
Let 2^x = y 2 ^x = 2 ² 2 ^x = 2 ¹
Then y² – 6y + 8 = 0 x = 2 x = 1    
Then factorize x = 1 and 2
y ² – 4y – 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
    
b)    5^2x + 4 x 5^x+1 – 125 = 0
    (5 ^x) ² + 4 x (5 ^x x 5¹) – 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, Factorize 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
5^x = -25
        
c)    3 ^2x – 9 = 0
    (3 ^x) ² – 9 = 0
    Let 3^x = a
     a² – 9 = 0
    a² = 9
    a = ±√9
    a = ± 3
    a = 3 or – 3
Since 3^x = a,     when a = 3
    3 ^x = 3¹
    x = 1
Since 3^x = a,     when a = -3
    3 ^x = – 3
    

 Evaluation:
Solve: (a) 3(2^{2x + 3}) – 5(2^x+2) – 156 = 0 (b )    9^2x+1 = (81 ^x-2/3^x)

 General Evaluation
Solve the following exponential equations.
a)    2<sup>2x
+ 1</sup> – 5 (2^x) + 2 = 0
b)    3^2x – 4 (3^x+1) + 27 = 0
    
Reading Assignment: Further Mathematics 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 (c) 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) 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