Re: DSE 2013 Paper Discussion

Posted by Nupur on May 20, 2014; 9:04am
URL: http://discussion-forum.276.s1.nabble.com/DSE-2013-Paper-Discussion-tp7584703p7590249.html

1(1+1.5/100)^t=2
where t is the doubling time (which we have to find out!!)
(1.015)^t=2
Taking log on both sides, get:
t log(1.015)= log 2
=>t = (log2/log 1.015)~(log 2)/0.015 which is the answer (option c)!
(Note: It's natural log everywhere!)