The equation `log_4(2-x)+log_(0.25)(2+x)=log_4(1-x)+log_(0.25)(2x+1)` has

Options

A. only one prime solution
B. two real solutions
C. no real solution
D. none of these

Answer

D

Explanation

`log_(4)(3-x)+log_(0.25)(3+x)=log_(4)(1-x)+log_(0.25)(2x+1)`
` rArr log_(4)(3-x)-log_(4)(3+x)=log_(4)(1-x)-log_(4)(2x+1)`
` rArr log_(4)(3-x)+log_(4)(2x+1)=log_(4) (1-x)+log_(4)(3+x)`
` rArr (3-x)(2x+1)=(1-x)(3+x)`
` rArr 3+5x-2x^(2)=3 - 2x-x^(2)`
` rArr x^(2) - 7x = 0`
`rArr x = 0, 7`
Only x = 0 is solution and x = 7 is to be rejected.