Question

The value of $\frac{lo{g}_{3}54}{lo{g}_{486}3}-\frac{lo{g}_{3}1458}{lo{g}_{18}3}$ is:

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

Let ${\log }_{3}18=a$
$\therefore \frac{{\log }_{3}54}{{\log }_{486}3}-\frac{{\log }_{3}1458}{{\log }_{18}3}$

$=\frac{{\log }_{3}54}{\frac{\log 3}{\log 486}}-\frac{{\log }_{3}1458}{\frac{\log 3}{\log 18}}$

$={\log }_{3}54\times \frac{\log 486}{\log 3}-{\log }_{3}1458\times \frac{\log 18}{\log 3}$

$={\log }_{3}54\times {\log }_{3}486-{\log }_{3}1458\times {\log }_{3}18$
$={\log }_3(18\times 3){\log }_3(18\times 27)-{\log }_3(18\times 81){\log }_{3}18$
$=({\log }_{3}18+{\log }_{3}3)({\log }_{3}18+3{\log }_{3}3)-({\log }_{3}18+4{\log }_{3}3){\log }_{3}18$
$=(a+1)(a+3)-(a+4)a$
$=a^2+3+4a-a^2-4a$
$=3$