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The equation (log)(x+1)(x- .5)=(log)(x-0...

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

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`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`
...
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