hopeless404 wrote: Evaluate the following sum $$\frac{1}{\log_2{\frac{1}{7}}}+\frac{1}{\log_3{\frac{1}{7}}}+\frac{1}{\log_4{\frac{1}{7}}}+\frac{1}{\log_5{\frac{1}{7}}}+\frac{1}{\log_6{\frac{1}{7}}}-\frac{1}{\log_7{\frac{1}{7}}}-\frac{1}{\log_8{\frac{1}{7}}}-\frac{1}{\log_9{\frac{1}{7}}}-\frac{1}{\log_{10}{\frac{1}{7}}}$$ Just use $\log_a b \cdot \log_b a = 1$. The sum is: $$\log_{\frac{1}{7}} \frac{2\cdot 3\cdot 4\cdot 5 \cdot 6}{7 \cdot 8 \cdot 9 \cdot 10} = 1$$