Question

if $a=\frac{4}{3}{\cot }^2{30}^{\circ }+3{\sin }^2{60}^{\circ }-3{\sec }^2{30}^{\circ }-\frac{3}{4}{\tan }^2{30}^{\circ }$ and $b=3{\tan }^2{45}^{\circ }+\cos 0^{\circ }-\cot {90}^{\circ }$ then ${\log }_b\left(a\right)=$

• A. 2
• B. -1
• C. $\frac{1}{2}$
• D. $\frac{-1}{2}$

Answer 1

The correct option is C $\frac{1}{2}$

$a=\frac{4}{3}{\cot }^2{30}^{\circ }+3{\sin }^2{60}^{\circ }-3\cos {\sec }^2{60}^{\circ }-\frac{3}{4}{\tan }^2{30}^{\circ }$
$a=\frac{4}{3}(\sqrt{3}{)}^2+3(\frac{\sqrt{3}}{2}{)}^2-3(\frac{2}{\sqrt{3}}{)}^2-\frac{3}{4}(\frac{1}{\sqrt{3}}{)}^2$
$a=4+\frac{9}{4}-4+\frac{1}{4}$
$a=\frac{5}{2}$
$b=3{\tan }^2{45}^{\circ }+\cos 0^{\circ }-\cot {90}^{\circ }$
$b=3(1{)}^2+1-0$
$b=4$
then ${\log }_b\left(a\right)$
= $\frac{\log a}{\log b}$
= $\frac{\log \frac{5}{2}}{\log 4}$
= $\frac{\log 5-\log 2}{2\log 2}$