Question

If $x={\log }_{3}5,y={\log }_{17}25$, which one of the following is correct?

• A. $x<y$
• B. $x=y$
• C. $x>y$
• D. None of the above

Answer 1

The correct option is C $x>y$

$x={\log }_{3}5$ and $y={\log }_{17}25$

$\therefore x=\frac{{\log }_{10}5}{{\log }_{10}3}$ and $y=\frac{{\log }_{10}25}{{\log }_{10}17}$ $(\because {\log }_{a}b=\frac{{\log }_{c}b}{{\log }_{c}a})$

$y=\frac{2{\log }_{10}5}{2{\log }_{10}\sqrt{17}}$ $(\because {\log }_a{b}^m={\log }_{a}b)$

$y=\frac{{\log }_{10}5}{{\log }_{10}\sqrt{17}}$

${\log }_{10}3<{\log }_{10}\sqrt{17}(\because 3<\sqrt{17})$

$\Rightarrow \frac{1}{{\log }_{10}3}>\frac{1}{{\log }_{10}\sqrt{17}}$

$\Rightarrow \frac{{\log }_{10}5}{{\log }_{10}3}>\frac{{\log }_{10}5}{{\log }_{10}\sqrt{17}}$

$\Rightarrow x>y$