Law of logarithm

$\large \sum _{ n=1 }^{ 99 }{ \log_{10} { ( \frac { n+1 }{ n } ) } }\\ =\log_{10} \prod_{n=1}^{99}\dfrac{n+1}{n}$ $\large =\log_{10} \dfrac{\not 2}{1}\times\dfrac{\not 3}{\not 2}\times\dfrac{\not 4}{\not 3}...\times\dfrac{100}{\not 99}$ $\large =\log_{10} 100=\huge\boxed{\color{#007fff}{2}}$

$\sum _{ n=1 }^{ 99 }{ \log { \left( \frac { n+1 }{ n } \right) } }= log(n+1)-log(n)$ $~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~ (\color{#302B94}{\small{ \text{Telescopic serie}}})$ $=log(2)-log(1)+log(3)-log(2)+log(4)-log(3)...log(100)-log(99)$ $=log(100)-log(1)=log({10}^{2})-log(1)=2log(10)-log(1)=\Large{\boxed{\color{#3D99F6}{2}}}$

Use telescopic rule..😁