How to solve $2^k - (bk + a) \ge 0 $?

Question

With given $a , b \in \mathbb{N} $, is there any way to find the smallest $ k \in \mathbb{N} $ that satisfies the following inequality (without trying for $k=1,2,...$):

$$2^k - (bk + a) \ge 0 $$

You say "inequation" (inequality?), but you have got equality there. — David, Jul 25 '13 at 07:47

score 3 · Answer 1 · edited Jul 25 '13 at 08:07

3

You can study the function $(0,\infty)\ni x\mapsto 2^x-(bx+a)$. Its derivative is $x\mapsto 2^x \ln 2 -b$. Its derivative is positive on... and negative on...

edited Jul 25 '13 at 08:07

answered Jul 25 '13 at 07:58

user37238

4,017

How to solve $2^k - (bk + a) \ge 0 $?

1 Answers1