Question

Find the value of $lo{g}_{\sqrt{3}}405-lo{g}_{\sqrt{3}}135+lo{g}_{3}9$

Answer 1

Given $lo{g}_{\sqrt{3}}405-lo{g}_{\sqrt{3}}135+lo{g}_{3}9$
We know that log a - log b = log ($\frac{a}{b}$)
And ${\log }_{3}9=2$
Simplifying the equation we get:-
=>$\log {}_{\sqrt{3}}\left(\frac{405}{135}\right)+2$
=> $lo{g}_{\sqrt{3}}3+2$
=> ${\log }_{\sqrt{3}}(\surd 3{)}^2+2$
=> $2lo{g}_{\sqrt{3}}\left(\surd 3\right)+2$( by applying power law)
=> 2 + 2
Answer : $lo{g}_{\sqrt{3}}405-lo{g}_{\sqrt{3}}135+lo{g}_{3}9=4$