Question

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

Answer 1

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