Logarithms are easy

$\begin{aligned} \log_3\left(1+\frac{1}{3}\right) + \log_3\left(1+\frac{1}{4}\right) + \log_3\left(1+\frac{1}{5}\right) + \cdots + \log_3\left(1+\frac{1}{80}\right) &= \log_3\left(\frac{4}{3}\right) + \log_3\left(\frac{5}{4}\right) + \log_3\left(\frac{6}{5}\right) + \cdots + \log_3\left(\frac{81}{80}\right) \\ &= \log_3\left(\frac{4}{3}\times\frac{5}{4}\times\frac{6}{5}\times\cdots\times\frac{81}{80}\right) \\ &= \log_3\left(\frac{81}{3}\right) = \log_3 27 = \boxed{3} \end{aligned}$

The given sum can be written as:

$\begin{aligned} S & = \sum_{k=3}^{80} \log_3 \left(1+\frac 1k\right) \\ & = \sum_{k=3}^{80} \log_3 \left(\frac {k+1}k \right) \\ & = \sum_{k=3}^{80} \big(\log_3(k+1)-\log_3 k \big) \\ & = \log_3 81 - \log_3 3 \\ & = 4-1= \boxed 3 \end{aligned}$