Solve the equation (logarithms)

Question

(i) Solve the equation log₂(x³+1) - 2 log₂x = log₂(x²-x+1) -2

(ii) Hence solve the equation log₂(x³+1) - log₂x² =log₂(x²-x+1)-2

 

 

Ans: (i) 2+2√2

       (ii) 2+2√2 , 2-2√2

Answer 1

(i) log₂(x³+1) - 2 log₂x = log₂(x²-x+1) -2

log₂ ((x³+1)/(x²(x²-x+1))) = -2

Now use long division to factor (x³+1) as (x+1)(x²-x+1):

log₂ ((x+1)/x²) = -2

(x+1)/x² = 2^-2

x+1 = (1/4)x²

x² - 4x - 4 = 0

 

Solve this quadratic equation, get x=2±2√2. Because of the term log₂x in the original equation, x has to be positive, so the only solution is x = 2+2√2.

 

(ii) Same as (i), only now with log₂x² in the original equation, x can be positive or negative, so the two solutions are x = 2±2√2.

Answer 2

Combine logarithms in the left side:

 

log₂(x³+1)-log₂x²-log₂(x²-x+1)=-2

 

Use the formula: log_ab+log_ac=log_abc;

 

log₂[(x³+1)/(x²(x²-x+1))]=-2;

log₂[(x+1)/x²]=-2

(x+1)/x²=¼; x>-1 (otherwise, log function is not defined, since argument of logarithm must be positive!)

 

x²-4x-4=0;

 

x₁=2-√8=2-2√2

x₂=2+2√2;

x₁>-1, so this root shall be kept;

 

Answer: 2+2√2; 2-2√2, so correct choice is (ii)