TITLE: approximation of $\frac{6\ln 10}{\ln 2}$?
QUESTION [5 upvotes]: The question is motivated from the following problem:

At a $15$ percent annual inflation rate, the value of the dollar would decrease by approximately one-half every $5$ years. At this inflation rate, in approximately how many years would the dollar be worth $\frac{1}{1,000,000}$ of its present value?

$A.25\quad B.50\quad C.75\quad D.100\quad E.125$
Finally one may get $10^6=2^n$ and $5n$ is the desired approximation. With a step of calculation, 
$$n=\frac{6\ln10}{\ln 2}.$$
I have no idea how to go on without a calculator. Are there any tricks/techniques for solving the problem?

REPLY [3 votes]: Since $6 \ln 10 = \ln(10^6)$, the question asks: $10^6$ is approximately what power of 2?  
You probably know that in science, "kilo" means 1000, but on computers a kilobyte is 1024.  Why the strange number?  Because it is a power of 2, and computers like powers of 2.  So remember this: $10^3 \approx 2^{10}$.  For your problem, square to get $10^6 \approx 2^{20}$.  So your answer is approximately $20$.