Logarithm of 1 (Special case)

For any base $b > 0$, we have $\log_b(1) = 0$. Why? Because $b^0 = 1$ for any $b$.

This means $\log_2(1) = 0$, $\log_{10}(1) = 0$, and $\ln(1) = 0$. You raise anything to the power $0$ and get $1$, so the logarithm of $1$ is always $0$.

This comes up in base cases when analyzing recursive algorithms. When your input size is $1$, the logarithm contributes $0$ to the complexity.