Question

Which of the following when simplified, reduces to zero?

• A. $\frac{1}{{\log }_{3}2}+\frac{2}{{\log }_{9}4}-\frac{3}{{\log }_{27}8}$
• B. ${\log }_2\left(\frac{2}{3}\right)+{\log }_4\left(\frac{9}{4}\right)$
• C. $-{\log }_8{\log }_4{\log }_{2}16$
• D. ${\log }_{10}\cot 1^{\circ }+{\log }_{10}\cot 2^{\circ }+{\log }_{10}\cot 3^{\circ }+........+{\log }_{10}\cot {89}^{\circ }$

Answer 1

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

The correct options are
A $\frac{1}{{\log }_{3}2}+\frac{2}{{\log }_{9}4}-\frac{3}{{\log }_{27}8}$
B ${\log }_2\left(\frac{2}{3}\right)+{\log }_4\left(\frac{9}{4}\right)$
C $-{\log }_8{\log }_4{\log }_{2}16$
D ${\log }_{10}\cot 1^{\circ }+{\log }_{10}\cot 2^{\circ }+{\log }_{10}\cot 3^{\circ }+........+{\log }_{10}\cot {89}^{\circ }$

$1)\frac{1}{{\log }_3\left(2\right)}+\frac{2}{{\log }_9\left(4\right)}-\frac{3}{{\log }_{27}\left(8\right)}$

$=\frac{\ln \left(3\right)}{\ln \left(2\right)}+\frac{2\ln \left(9\right)}{\ln 4}-\frac{3\ln 27}{\ln 8}$

$=\frac{\ln \left(3\right)}{\ln \left(2\right)}+\frac{2\ln \left(3^2\right)}{\ln 2^2}-\frac{3\ln 3^3}{\ln 2^3}$

$=\frac{\ln \left(3\right)}{\ln \left(2\right)}+\frac{2.2\ln \left(3\right)}{2.\ln 2}-\frac{3.3\ln 3}{3.\ln 2}$

$=0$.

$2\left){\log }_2\right(\frac{2}{3})+\frac{\ln \frac{9}{4}}{\ln 4}$

$={\log }_2\left(\frac{2}{3}\right)+\frac{\ln \frac{9}{4}}{2\ln 2}$

$={\log }_2\left(\frac{2}{3}\right)+\frac{1}{2}{\log }_2\left(\frac{9}{4}\right)$

$={\log }_2\left(\frac{2}{3}\right)+{\log }_2\left(\frac{3}{2}\right)$

$={\log }_2(\frac{2}{3}\times \frac{3}{2})$

$={\log }_2\left(1\right)=0$

$3)-{\log }_8({\log }_4\left({\log }_2\right(16\left)\right))$

$=-{\log }_8\left({\log }_4\right(4\left)\right)$

$=-{\log }_8\left(1\right)$

$=0$

$4){\log }_{10}cot{1}^0+{\log }_{10}cot{2}^0+....{\log }_{10}cot({89}^0)$

$={\log }_{10}(cot{1}^0.cot{2}^0...cot{89}^0)$

$={\log }_{10}\left[\right(cot{1}^0.cot{89}^0).(cot{2}^0.cot{88}^0)...co{t}^2{45}^0)]$

$={\log }_{10}[cot{1}^0.tan{1}^0).(cot{2}^0.tan{2}^0)...co{t}^2{45}^0\left)\right]$

$={\log }_{10}\left(1\right)$

$=0$.