Question

Which of the following, when simplified reduces to unity?

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

The correct options are
A ${\log }_{10}5.{\log }_{10}20+{\left({\log }_{10}2\right)}^2$
B $\frac{2\log 2+\log 3}{\log 48-\log 4}$
C $-{\log }_5{\log }_3\sqrt{\sqrt[5]{59}}$

Let $a={\log }_{10}5.{\log }_{10}20+{\left({\log }_{10}2\right)}^2$

$\Rightarrow a=\frac{\left(\log 5\right)\left(\log 20\right)}{{\left(\log 10\right)}^2}+{\left(\frac{\log 2}{\log 10}\right)}^2$

$\Rightarrow a=\frac{\left(\log 5\right)\left(\log (2^2\times 5)\right)}{{\left(\log 10\right)}^2}+{\left(\frac{\log 2}{\log 10}\right)}^2$

$\Rightarrow a=\frac{\log 5(2\log 2+\log 5)+{\left(\log 2\right)}^2}{{(\log 2+\log 5)}^2}=\frac{{\left(\log 2\right)}^2+2\log 5\log 2+{\left(\log 5\right)}^2}{{(\log 2+\log 5)}^2}$

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

$\Rightarrow a=1$

Let $b=\frac{2\log 2+\log 3}{\log 48-\log 4}$

$\Rightarrow b=\frac{\log (2^2\times 3)}{\log \left(\frac{48}{4}\right)}=\frac{\log 12}{\log 12}=1$

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

$\Rightarrow c=-{\log }_5\left(\frac{{\log }_3\sqrt[\frac{5}{2}]{\frac{5}{2}3}}{2}\right)=-{\log }_5\left(\frac{2{\log }_{3}3}{10}\right)$

$\Rightarrow c=-{\log }_5\left(\frac{1}{5}\right)=1$

Let $d=\frac{1}{6}{\log }_{\sqrt{3}/2}\left(\frac{64}{27}\right)$
$\Rightarrow d=\frac{1}{6}{\log }_{\sqrt{3}/2}{\left(\frac{4}{3}\right)}^3$

$\Rightarrow d={\log }_{\sqrt{3}/2}{\left(\frac{4}{3}\right)}^{\frac{3}{6}}$

$\Rightarrow d={\log }_{\sqrt{3}/2}\left(\frac{2}{\sqrt{3}}\right)=-1\ne 1$

$[\because \log a^m=m\log a,{\log }_{a}a=1]$
Ans: A,B,C