Question

If ${\log }_{3}2=x$, then the value of $\frac{{\log }_{10}72}{{\log }_{10}24}$ is:

• A. $\frac{1+x}{1-x}$
• B. $\frac{2+3x}{1+3x}$
• C. $\frac{2-3x}{2+3x}$
• D. $\frac{3x+1}{3x+2}$

Answer 1

Answer: (B)

Given, ${\log }_{3}2=x$
$\frac{{\log }_{10}72}{{\log }_{10}24}={\log }_{24}72={\log }_{24}(24\times 3)$
$={\log }_{24}24+{\log }_{24}3=1+\frac{\log 3}{\log 24}$
$=1+\frac{{\log }_{3}3}{{\log }_{3}24}$
$=1+\frac{1}{{\log }_3(3\times 8)}$
$=1+\frac{1}{{\log }_{3}3+{\log }_{3}8}$
$=1+\frac{1}{1+{\log }_3{2}^3}$
$=1+\frac{1}{1+3{\log }_{3}2}$
$=\frac{1+3{\log }_{3}2+1}{1+3{\log }_{3}2}=\frac{2+3x}{1+3x}.$