Question

The number of real solution/s of the equation $9^{{\log }_3\left({\log }_e^x\right)}={\log }_{e}x-({\log }_{e}x{)}^2+1$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

2

$9^{{\log }_3\left({\log }_{e}x\right)}={\log }_{e}x-{\left({\log }_{e}x\right)}^2+1$
$3^{2{\log }_3\left({\log }_{e}x\right)}={\log }_{e}x-{\left({\log }_{e}x\right)}^2+1$
$3^{{\log }_3{\left({\log }_{e}x\right)}^2}={\log }_{e}x-{\left({\log }_{e}x\right)}^2+1$
${\left({\log }_{e}x\right)}^2={\log }_{e}x-{\left({\log }_{e}x\right)}^2+1$
$2{\left({\log }_{e}x\right)}^2-{\log }_{e}x-1=0$
$2{\left({\log }_{e}x\right)}^2-2{\log }_{e}x+{\log }_{e}x-1=0$
$2{\log }_{e}x[{\log }_{e}x-1]+1[{\log }_{e}x-1]=0$
$[{\log }_{e}x-1][2{\log }_{e}x+1]=0$
$\therefore {\log }_{e}x=1$ and ${\log }_{e}x=\frac{-1}{2}$
$e^{{\log }_{e}x}=e^1$ and $e^{{\log }_{e}x}=e^{\frac{-1}{2}}$
$x=e$ and $x=\frac{1}{\sqrt{e}}$
$\therefore$ 2 Real solutions