Adding Almost One Logs

Relevant wiki: Properties of Logarithms - Basic

By the Laws of Logarithm: $\large\color{#3D99F6}{\log MN = \log M + \log N}$ , so....

$\begin{aligned} & \large\log \frac{1}{2} + \log \frac{2}{3} + \log \frac{3}{4} + \cdots + \log \frac{99}{100} \\ \implies & \large\log\bigg( \frac{1}{2} * \frac{2}{3} * \frac{3}{4} * \cdots * \frac{99}{100} \bigg) \\ \implies & \large\log\bigg(\frac{1}{\color{#D61F06}{\cancel{\color{#333333}{2}}}} * \frac{\color{#D61F06}{\cancel{\color{#333333}{2}}}}{\color{#20A900}{\cancel{\color{#333333}{3}}}} * \frac{\color{#20A900}{\cancel{\color{#333333}{3}}}}{\color{#302B94}{\cancel{\color{#333333}{4}}}} * \cdots * \frac{\color{#EC7300}{\cancel{\color{#333333}{99}}}}{100}\bigg) \\ \implies & \large\log\frac{1}{100} \\ \implies & \large\log {10^{-2}} = \color{#3D99F6}{\boxed{-2}} \end{aligned}$

$\log \frac{1}{2} + \log \frac{2}{3} + \log \frac{3}{4} + \cdots + \log \frac{99}{100}=$

using properties of logs: $\color{#D61F06}{\log \frac{a}{b}=\log a - \log b }$ , we obtain

$(\log1-\log 2)+(\log 2-\log 3)+(\log 3-\log 4)+\log 4+...+(\log 99+\log 100)=$

canceling few terms we will end up with:

$\log(1)-\log(100)=\log 1- \log 10^2 =$

using another property of logs which is $\color{#3D99F6}{\log a^r = r \log ( a)}$ and knowing $\color{#3D99F6}{\log 1= 0}$ gives us:

$\ 0-2\log(10)= \boxed{-2}$