Prove that for any 9-digit number (base 10) there is an integral power of 2 whose first 9 digits are that...

arithmetic/digits/power.two

Problem

Prove that for any 9-digit number (base 10) there is an integral power
of 2 whose first 9 digits are that number.

Show solution

Let v = log to base 10 of 2.
Then v is irrational.

Let w = log to base 10 of these 9 digits.

Since v is irrational, given epsilon > 0, there exists some natural number 
n such that

   {w} < {nv} < {w} + epsilon

({x} is the fractional part of x.)  Let us pick n for when 

   epsilon = log 1.00000000000000000000001.

Then 2^n does the job.