Question

The equation `(log)_(x+1)(x- .5)=(log)_(x-0. 5)(x+1)` has (A) two real solutions (B) no prime solution (C) one integral solution (D) no irrational solution

Answer 1

`log_(x+1)(x-0.5)=log_(x-0.5)(x+1)`
`(log_2(x-0.5))/(log_2(x+1))=(log_2(x+1))/(log(x-0.5))`
`(log(x-0.5)]^2-[log(x+1)]^2=0`
`[log(x-0.5)+log(x+1)][log(x-0.5)-log(x+1)]=0`
`log((x-0.5)/(x+1))=0`
`(x-0.5)/(x+1)=1`
`x-0.5=x+1`
`log(x-0.5)(x+1)=0`
`(x-0.5)(x+1)=1`
`(x-1/2)(x+1)=1`
`(2x-1)(x+1)=2`
`2x^2+2x-x-1-2=0`
`2x^2+x-3=0`
`2x^2+3x-2x-3=0`
`x(2x+3)-1(2x+3)=0`
`(x-1)(2x+3)=0`
`x=1,-3/2`
x can not be`-3/2`
Option B,C and D are correct.