Question

The value of ${\log }_7\left({\log }_7\sqrt{7\left(\sqrt{7\sqrt{7}}\right)}\right)$ is:

• A. $1-3{\log }_{7}2$
• B. $3{\log }_{7}2$
• C. $7-3{\log }_{7}2$
• D. $1-2{\log }_{7}3$

Answer 1

The correct option is A $1-3{\log }_{7}2$

Let, $L={\log }_7\left({\log }_7\sqrt{7\left(\sqrt{7\sqrt{7}}\right)}\right)$

Let's simplify $\sqrt{7\left(\sqrt{7\sqrt{7}}\right)}$ first.

$\sqrt{7\left(\sqrt{7\sqrt{7}}\right)}=\sqrt{7\left(\sqrt{7\times 7^{1/2}}\right)}$
$=\sqrt{7\left(\sqrt{7^{3/2}}\right)}$
$=\sqrt{7{\left(7^{3/2}\right)}^{1/2}}$
$=\sqrt{7\times 7^{3/4}}$
$=\sqrt{7^{7/4}}$
$={\left(7^{7/4}\right)}^{1/2}$
$=7^{7/8}$

$\therefore L={\log }_7\left({\log }_7\sqrt{7\left(\sqrt{7\sqrt{7}}\right)}\right)$
$={\log }_7\left({\log }_7{7}^{7/8}\right)$
$={\log }_7\left(\frac{7}{8}\right)$
$={\log }_{7}7-{\log }_{7}8=1-{\log }_7{2}^3=1-3{\log }_{7}2$