Question

${\log }_{10}(1+\frac{1}{2})+{\log }_{10}(1+\frac{1}{3})+{\log }_{10}(1+\frac{1}{4})+...+{\log }_{10}(1+\frac{1}{1999})$ When simplified has the value equal to

• A. 1
• B. 3
• C. 10
• D. 100

Answer 1

The correct option is B 3

Using
${\log }_{10}a$ + ${\log }_{10}b$ = ${\log }_{10}ab$
= ${\log }_{10}(\frac{3}{2}\times \frac{4}{3}\times \frac{5}{4}\times .......\frac{2000}{1999})$
= ${\log }_{10}\left(\frac{2000}{2}\right)$..... remaining terms cancel with each other
= ${\log }_{10}1000$
$=3$