Question

Solve the following:
$\cos \left(\frac{\pi }{10}({\log }_3\frac{1}{9}+{\log }_{\frac{1}{9}}3)\right)$

Answer 1

$lo{g}_3\frac{1}{9}=x_1$

$\frac{1}{9}=3^{x_1}$

$3^{-2}=3^{x_1}$

Now compare only with the powers

$x_1=-2$

Now, $lo{g}_{\frac{1}{9}}3=x_2$

$(\frac{1}{9}{)}^{x_2}=3$

$(3{)}^{-2x_2}=3^1$

now compare with the powers we get,

$x_2=-\frac{1}{2}$

Now put the both value to the given expression,

$=\cos \left(\frac{\pi }{10}\right(-2-\frac{1}{2}\left)\right)$

$=\cos \left(\frac{\pi }{10}\right(\frac{-5}{2})$

$=\cos \left(\frac{-\pi }{4}\right)$

$=\cos \left(\frac{\pi }{4}\right)$

$=\frac{1}{\sqrt{2}}$