Question

The solution of the equation $4^{{\log }_2\log x}=\log x-{\left(\log x\right)}^2+1$ is

• A. $x=1$
• B. $x=4$
• C. $x=e$
• D. $x=e^2$

Answer 1

Answer: (C)

$x=e$

${\log }_2\log x$ is meaningful if $x>1$
Since $4^{{\log }_2\log x}={\left(2^{{\log }_2\log x}\right)}^2={\left(\log x\right)}^2$ .......$[\because a^{{\log }_{a}x}=x,a>0,a\ne 1]$
So the given equation {$4^{{\log }_2\log x}=\log x-{\left(\log x\right)}^2+1$} reduces to
$2{\left(\log x\right)}^2-\log x-1=0$
$\Rightarrow \log x=1,\log x=-\frac{1}{2}$
But for $x>1$, $\log x>0$
so $\log x=1$ i.e., $x=e$
Ans: C