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Mathematics

Logarithmic Equation

Solve the following equation

  1. a) Log10 (x2 – 4x + 7) = 2
  2. b) Log8 (r2 – 8r + 18) = 1/3

Solution

  1. a) Log10 (x2 – 4x + 7) = 2

X2 – 4x + 7 = 102 (index form)

X2 – 4x + 7 = 100

X2 – 4x + 7 – 100 = 0

X2 – 4x – 93 = 0

Using quadratic formular

= – b ±√b2– 4ac

2a

a = 1, b = -4, c = – 93

x = – (- 4) ± √(- 4) – 4 X 1 X (- 93)

2 X 1

= + 4 ± √16 + 372

2

= + 4 ± √388/2

= x = 4 +√ 388/2 or 4 – √388/2

x =  11.84 or x = – 7.85

2)            Log8 (x2 – 8x + 18) =81/3

X2 – 8x + 18 = 81/3

X2 – 8x + 18 = (2)3X1/3

X2 – 8x + 18 =2

X2 – 8x 18 – 2 = 0

X2 – 8x + 16 = 0

X2 – 4x – 4x + 16 = 0

X(x – 4) -4 (x – 4) = 0

(x – 4) (x – 4) = 0

(x – 4) twice

X = + 4 twice

Change of Base

Let logbP = x and this means P = bx

LogcP = logcbx = x logcb

If x logcb = logcP

X = logcP

Logc b

:.             LogcP = logcP

Logcb

Example :  Shows that logab X logba = 1

Logab = logcb

Logca

Logba = logca

Logcb

:.             Logab X logba = logcb  X logca

Logca  +  logcb= 1

EVALUATION ( USE THE DISCUSSION BOX AT THE BOTTOM TO SUBMIT YOUR ANSWER FOR DISCUSSION AND APPRAISAL)

Solve the following logarithm equation.

Log3 (x2 + 7x + 21) = 2

Log10 (x2 – 3x + 12) = 1

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