Question

$\frac{lo{g}_{8}17}{lo{g}_{9}23}-\frac{lo{g}_{2\sqrt{2}}17}{lo{g}_{3}23}$ is equal to

• A. 0
• B. 1
• C. $\frac{17}{8}$
• D. $\frac{23}{17}$

Answer 1

The correct option is A

0

$\frac{lo{g}_{8}17}{lo{g}_{9}23}-\frac{lo{g}_{2\sqrt{2}}17}{lo{g}_{3}23}$

$=\frac{\frac{lo{g}_{2}17}{lo{g}_2{2}^3}}{\frac{lo{g}_{3}23}{lo{g}_3{3}^2}}-\frac{\frac{lo{g}_{2}17}{lo{g}_2{2}^{3/2}}}{\frac{lo{g}_{3}23}{lo{g}_{3}3}}$

$=\frac{\frac{1}{3}lo{g}_{2}17}{\frac{1}{2}lo{g}_{3}23}-\frac{\left(\frac{2}{3}\right)lo{g}_{2}17}{lo{g}_{3}23}$

$=\frac{2}{3}\frac{lo{g}_{2}17}{lo{g}_{3}23}-\frac{2}{3}\frac{lo{g}_{2}17}{lo{g}_{3}23}$

$=0$