Question

If $4^{{\log }_2\log x}=\log x-{\left(\log x\right)}^2+1$ (base is e), then find the value of $x$

• A. $x=e$
• B. $x=2e$
• C. $x=3e$
• D. none of these

Answer 1

The correct option is A $x=e$

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