The number of solutions of the equation `log_(x+1)(x-0.5)=log_(x-0.5)(x+1)` is

Options

A. two real solutions
B. no prime solution
C. one integral solution
D. no irrational solution

Answer

B, C, D

Explanation

`(log_(2)(x-0.5))/(log_(2)(x+1)) = (log_(2)(x+1))/(log_(2)(x-0.5))`
` or [log_(2)(x+1)]^(2)=[log_(2)(x-0.5)]^(2)`
` :. log_(2)(x+1)=pmlog_(2)(x-0.5)`
If ` log_(2)(x+1)=log_(2)(x-0.5)`.Then,
` x + 1 = x - 0.5 `, hence no solution
If ` log_(2)(x+1)= log(x-0.5)^(-1)`. Then,
` x+ 1 = 1/(x-(1//2)) = 2/(2x-1)`
` or (x+1)(2x-1)=2`
` or 2x^(2) + x - 3 = 0`
` or 2x^(2) + 3x - 2x - 3 = 0`
` or (x-1)(2x+3) = 0`
`rArr x = 1 " "(x =- 3//2" is rejected" )`