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) ² – 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) =8^1/3

X² – 8x + 18 = 8^1/3

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 log_bP = x and this means P = b^x

Log_cP = log_cb^x = x log_cb

If x log_cb = log_cP

X = log_cP

Log_c b

:.             Log_cP = log_cP

Log_cb

Example :  Shows that log_ab X log_ba = 1

Log_ab = log_cb

Log_ca

Log_ba = log_ca

Log_cb

:.             Log_ab X log_ba = log_cb  X log_ca

Log_ca  +  log_cb= 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