Question

The value of $5^{{\log }_{\frac{1}{5}}\left(\frac{1}{2}\right)}+{\log }_{\sqrt{2}}\left(\frac{4}{\sqrt{3}+\sqrt{7}}\right)+{\log }_{\frac{1}{2}}\left(\frac{1}{10+2\sqrt{21}}\right)$ is

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is D $6$

$5^{{\log }_{\frac{1}{5}}\left(\frac{1}{2}\right)}=5^{-{\log }_5\left(\frac{1}{2}\right)}=5^{{\text{log}}_{5}2}=2(\because a^{{\log }_{a}b}=b)$

${\log }_{\sqrt{2}}\left(\frac{4}{\sqrt{3}+\sqrt{7}}\right)={\log }_{\sqrt{2}}(\sqrt{7}-\sqrt{3})={\log }_2(10-2\sqrt{21})$

${\log }_{\frac{1}{2}}\left(\frac{1}{10+2\sqrt{21}}\right)={\log }_2(10+2\sqrt{21})$

$\therefore 5^{{\log }_{\frac{1}{5}}\left(\frac{1}{2}\right)}+{\log }_{\sqrt{2}}\left(\frac{4}{\sqrt{3}+\sqrt{7}}\right)+{\log }_{\frac{1}{2}}\left(\frac{1}{10+2\sqrt{21}}\right)$
$=2+{\log }_2(10-2\sqrt{21})+{\log }_2(10+2\sqrt{21})$
$=2+{\log }_{2}16$
$=2+4=6$