Question

The value of ${\log }_3(1+\frac{1}{3})+{\log }_3(1+\frac{1}{4})+\cdots +{\log }_3(1+\frac{1}{80})$ is

• A. $0$
• B. $1$
• C. $3$
• D. $6$

Answer 1

The correct option is C $3$

${\log }_3(1+\frac{1}{3})+{\log }_3(1+\frac{1}{4})+\cdots +{\log }_3(1+\frac{1}{80})$

$={\log }_3\left(\frac{4}{3}\right)+{\log }_3\left(\frac{5}{4}\right)+\cdots +{\log }_3\left(\frac{81}{80}\right)$

By applying the log multiplication rule, we get

$={\log }_3(\frac{4}{3}\times \frac{5}{4}\times \cdots \frac{81}{80})$

$={\log }_3\left(\frac{81}{3}\right)$

$={\log }_3\left(27\right)$

$\Rightarrow {\log }_3{\left(3\right)}^3=3{\log }_3\left(3\right)=3$

Hence, option $C$ is the correct answer.