Question

find the value of $\frac{{\log }_{4}7}{{\log }_{4}5}-\frac{{\log }_{9}5}{{\log }_{9}7}$

Answer 1

$\frac{{\log }_{4}7}{{\log }_{4}5}-\frac{{\log }_{9}5}{{\log }_{9}7}$

We know that $\frac{{\log }_{a}b}{{\log }_{a}c}={\log }_{c}b$

$\therefore \frac{{\log }_{4}7}{{\log }_{4}5}-\frac{{\log }_{9}5}{{\log }_{9}7}$$={\log }_{5}7-{\log }_{7}5$

We know that ${\log }_{b}a=\frac{1}{{\log }_{a}b}$

$\frac{{\log }_{4}7}{{\log }_{4}5}-\frac{{\log }_{9}5}{{\log }_{9}7}={\log }_{5}7-\frac{1}{{\log }_{5}7}$

$=\frac{{\left({\log }_{5}7\right)}^2-1}{{\log }_{5}7}$

Take $1={\left({\log }_{5}5\right)}^2$

$=\frac{{\left({\log }_{5}7\right)}^2-{\left({\log }_{5}5\right)}^2}{{\log }_{5}7}$

$=\frac{({\log }_{5}7+{\log }_{5}5)({\log }_{5}7-{\log }_{5}5)}{{\log }_{5}7}$

$=\frac{\left({\log }_{5}35\right)\left({\log }_5\frac{7}{5}\right)}{{\log }_{5}7}$

$\frac{{\log }_{4}7}{{\log }_{4}5}-\frac{{\log }_{9}5}{{\log }_{9}7}={\log }_{7}35{\log }_5\frac{7}{5}$