Question

If $\log x,\log \sqrt{6-2x},log(x-1)$ are in A.P. then, number of possible value of $x$ is/are

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

$x>0$ , $x>1$ and $6-2x>0$

As $logx,log\sqrt{6-2x},log(x-1)$ are in A.P
$2log\sqrt{6-2x}=logx+log(x-1)\Rightarrow 6-2x=x(x-1)\Rightarrow 6-2x=x^2-x\Rightarrow x^2+x-6=0\Rightarrow (x+3)(x-2)=0$
$\Rightarrow x=-3$ or $x=2$
AS $x\ne -3$ $\therefore x=2$