Question

Which of the following when simplified, reduces to unity?

• A. ${\log }_{10}5.{\log }_{10}20+({\log }_{10}2{)}^2$
• B. $\frac{2\log 2+\log 3}{\log 48-\log 4}$
• C. $-{\log }_5{\log }_3{\sqrt{9}}^{1/5}$
• D. $\frac{1}{6}{\log }_{\sqrt{\frac{3}{4}}}\left(\frac{64}{27}\right)$

Answer 1

Answer: (A), (B), (C)

The correct options are
A ${\log }_{10}5.{\log }_{10}20+({\log }_{10}2{)}^2$
B $\frac{2\log 2+\log 3}{\log 48-\log 4}$
C $-{\log }_5{\log }_3{\sqrt{9}}^{1/5}$

(A) ${\log }_{10}5.{\log }_{10}20+({\log }_{10}2{)}^2$

$={\log }_{10}5.{\log }_{10}(4\times 5)+({\log }_{10}2{)}^2$

$={\log }_{10}5(2{\log }_{10}2+{\log }_{10}5)+\left(\right({\log }_{10}2{)}^2)$

$={({\log }_{10}2+{\log }_{10}5)}^2$

$={\left({\log }_{10}10\right)}^2$

$=1$ ...$[\because {\log }_{a}a=1]$
(B) $\frac{2\log 2+\log 3}{\log 48-\log 4}$
$=\frac{2\log 2+\log 3}{\log (16\times 3)-\log 4}$
$=\frac{\log 4+\log 3}{2\log 4+\log 3-\log 4}$
$=\frac{\log 4+\log 3}{\log 4+\log 3}$

$=1$
(C) $-{\log }_5{\log }_3{\sqrt{9}}^{\frac{1}{5}}$

$=-{\log }_5{\log }_3{3}^{\frac{2}{10}}$

$=-{\log }_5{\log }_3{3}^{\frac{1}{5}}$

$=-{\log }_5\frac{1}{5}=1$
(D) $\frac{1}{6}{\log }_{\sqrt{\frac{3}{4}}}\frac{64}{27}$

$=\frac{3}{6}{\log }_{\sqrt{\frac{3}{4}}}\frac{4}{3}$

$=\frac{1}{2}{\log }_{\sqrt{\frac{3}{4}}}\frac{4}{3}$

$={\log }_{\frac{3}{4}}\frac{4}{3}=-1$