Question

If $\frac{1}{k}=\frac{3}{{\log }_2(3000{)}^7}+\frac{1}{{\log }_3(3000{)}^7}+\frac{3}{{\log }_5(3000{)}^7}$, then the value of $k$ is

Answer 1

$\frac{3}{{\log }_2(3000{)}^7}+\frac{1}{{\log }_3(3000{)}^7}+\frac{3}{{\log }_5(3000{)}^7}=\frac{3}{7{\log }_{2}3000}+\frac{1}{7{\log }_{3}3000}+\frac{3}{7{\log }_{5}3000}=\frac{1}{7}[3{\log }_{3000}2+{\log }_{3000}3+3{\log }_{3000}5]=\frac{1}{7}[3{\log }_{3000}10+{\log }_{3000}3]=\frac{1}{7}\left[{\log }_{3000}({10}^3\times 3)\right]=\frac{1}{7}\left[{\log }_{3000}3000\right]\Rightarrow \frac{1}{k}=\frac{1}{7}\therefore k=7$