Logs

Algebra Level 2

$\log\left(\dfrac{1}{2}\right) + \log\left(\dfrac{2}{3}\right) + \log\left(\dfrac{3}{4}\right)+ ...+ \log\left(\dfrac{99}{100}\right) = \ ?$

-2 -3 2 0 1

2 solutions

Omkar Kulkarni
Jan 25, 2015

The expression becomes

$\log_{}{\frac {1}{2} \times \frac {2}{3} \times \frac {3}{4} \times...\times \frac {99}{100}}$

$= \log_{}{\frac {1}{100}}$

$= \log_{}{1} - \log_{}{100}$

$= \log_{10}{1} - \log_{10}{100}$

$= 0-2 = \boxed {-2}$

Was the second last step right? I'm not sure; not very familiar with logarithms.

second last step is ok :)

Rhoy Omega - 6 years, 4 months ago

Oh okay. Thank you! :)

Omkar Kulkarni - 6 years, 4 months ago

Lu Chee Ket
Jan 25, 2015

Base 10. Log (1/ 100) = log (10^-2) = -2