Question

The value of ${\log }_4\sqrt{4\sqrt{4\sqrt{4\dots \infty }}}+\log$ $\left(\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3+\dots \infty \right)+\left({\log }_{10}2+{\log }_{10}5\right)$is___

Answer 1

Answer: 2

Given: ${\log }_4\sqrt{4\sqrt{4\sqrt{4\dots \infty }}}+\log (\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3+\dots \infty )$
Let $x=\sqrt{4x}$
$\Rightarrow x^2=4x\Rightarrow x^2-4x=0$
$\Rightarrow x(x-4)=0$
$\Rightarrow x=4(\because x\ne 0)$
$\therefore {\log }_4\sqrt{4\sqrt{4\sqrt{4\dots \infty }}}={\log }_{4}4=1$
Now, $\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3+\dots \infty$ is a G.P.
Here, $a=\frac{1}{2},r=\frac{1}{2}$
So, $s_{\infty }=\frac{a}{1-r}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1$
$\therefore \log \left(\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3+\dots \infty \right)=\log 1=0$
and, ${\log }_{10}2+{\log }_{10}5={\log }_{10}2.5=10=1$
Hence, ${\log }_4\sqrt{4\sqrt{4\sqrt{4\dots \infty }}}+\log (\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+$
${\left(\frac{1}{2}\right)}^3+\dots \infty )+{\log }_{10}2+{\log }_{10}5$
$=1+0+1=2$