Question

Which of the following, when simplified, reduces to unity ?

• A. ${\log }_{10}5\cdot {\log }_{10}20+({\log }_{10}2{)}^2$
• B. $\frac{2\log 2+\log 3}{\log 48-\log 4}$
• C. $-{\log }_5{\log }_3\sqrt{\sqrt[5]{59}}$
• D. $\frac{1}{6}{\log }_{\sqrt{3}/2}\left(\frac{64}{27}\right)$

Answer 1

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

$-{\log }_5{\log }_3\sqrt{\sqrt[5]{59}}$

${\log }_{10}5\cdot {\log }_{10}20+({\log }_{10}2{)}^2$
$={\log }_{10}\left(\frac{10}{2}\right)\cdot {\log }_{10}(10\times 2)+({\log }_{10}2{)}^2$
$=(1-{\log }_{10}2)(1+{\log }_{10}2)+({\log }_{10}2{)}^2$
$=1-({\log }_{10}2{)}^2+({\log }_{10}2{)}^2$
$=1$

$\frac{2\log 2+\log 3}{\log 48-\log 4}=\frac{\log (2^2\times 3)}{\log \left(48/4\right)}=\frac{\log 12}{\log 12}=1$

$-{\log }_5{\log }_3\sqrt{\sqrt[5]{59}}=-{\log }_5{\log }_3{9}^{1/10}$
$=-{\log }_5{\log }_3{3}^{1/5}$
$=-{\log }_5\left(1/5\right)=1$

$\frac{1}{6}{\log }_{\sqrt{3}/2}\left(\frac{64}{27}\right)=\frac{1}{6}{\log }_{\sqrt{3}/2}{\left(\frac{2}{\sqrt{3}}\right)}^6=-1$