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# INDICIAL AND EXPONENTIAL EQUATIONS

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

1. 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

1. a)(1/2) X = 8                                                     b)      (0.25) x+1 = 16

(2 -1x = 2 3                                                                (25/100) x+1 =  42

–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

1. c)3x = 1/81                                                               d) 10 x  = 1/0.001

3x = 1/34                                                                     10 x  = 1000

3x = 3 -4                                                                       10 x  = 10 3

x = -4                                                                         10 x  = 10 3

x = 3

1. e)4/2x = 64 x

4÷2x = 64 x

22 ÷2x = 64 x

2-x = (2 6x

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

1. a)2 x = 0.125      b) 25 (5x) = 625                   c)  10 x = 1/100000

Some exponential equation can be reduced to quadratic form as can be seen below.

Example:

Solve the following equations.

1. a)22x – 6 (2x) + 8 = 0
2. b)52x + 4 x 5 x+1 – 125 = 0
3. c)32x – 9 = 0

Solution

1. a)   22x – 6 (2x) + 8 = 0                                 When y = 4 then,         and         When y = 2 then,

(2x)2 – 6 (2x) + 8 = 0                                          2 x = 4                                       2 x = 2

Let 2x = y                                                          2 x = 2 2                                                          2 x = 2 1

Then y2 – 6y + 8 = 0                                           x = 2                                         x = 1

Then factorize                                                                     x = 1 and 2

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

1. b)52x + 4 x 5x+1 – 125 = 0

(5 x2 + 4 x (5 x x 51) – 125 = 0

Let 5 x = 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= p,           p = 5

5 = 5 1

x = 1

5x = -25

Evaluation

Solvethe following exponential equations

1. a)         2x= 0.125      b) 25 (5x) = 625                   c)  10x = 1/100000

Someexponential equation can be reduced to quadratic form as can be seen below.

Example:

Solvethe following equations.

1. a)         22x– 6 (2x) + 8= 0
2. b)         52x+ 4 x 5x+1 –125 = 0
3. c)          32x– 9 = 0

Solution

1. a)  22x– 6 (2x) + 8 = 0                                 When y = 4then,         and         When y = 2 then,

(2x)2– 6 (2x) + 8 = 0                                          2 x= 4                                      2 x = 2

Let 2x =y                                                         2 x = 2 2                                                         2 x = 2 1

Then y2 –6y + 8 = 0                                           x =2                                        x = 1

Then factorize                                                                    x = 1 and 2

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

1. b)         52x+ 4 x 5x+1–125 = 0

(5 x2 + 4 x(5 x x 51) – 125 = 0

Let 5 x = 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=p,           p = 5

5 = 5 1

x = 1

5x = -25

1. c)3 2x – 9 = 0

(3 x2 – 9 = 0

Let 3 = a

a2 – 9 = 0

a2 = 9

a = ±√9

a = ± 3

a = 3 or – 3

Since 3 = a,   when a = 3

3 x  = 31

x = 1

Since 3x = a,   when a = -3

3 x = – 3

Evaluation:

Solve:(a)   3(22x + 3) – 5(2x+2)- 156 = 0         (b )       92x+1 = (81 x-2/3x)

General Evaluation

Solvethe following exponential equations.

1. a)         22x+ 1– 5 (2x)+ 2 = 0
2. b)         32x– 4 (3x+1) +27 = 0

Reading AssignmentFurther Mathematics Project Book 1(New thirdedition).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 : 22x + 2x+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

Solvethe following exponential equations

(1) (3x)2 + 2(3x)– 3 = 0          (2) 52x+1 – 26(5x) + 5 = 0

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