44118 = 3^b mod 65537. Find b.

44118 = 3^b mod 65537

This says that

65537n + 44118 = 3^b

There are infinite solutions to this...for instance....

Let n=3

Then we have

240729 = 3^b

Taking the log of both sides, we have

log 240729 = log 3^b    .....and by a property of logs, we have

log 240729 = b*log3    ......and dividing both sides by log3, we have

log 240729/log 3  = b ≈ 11.2791630188882364

If we insist that both n and b be integers, I'm not sure how to solve this....(if it even has a solution).....(maybe Alan knows a method)

Take logs of both sides:  ln(44118) = ln(3^b)

From a property of logs we have ln(3^b) = b*ln(3) so ln(44118) = b*ln(3); hence b = ln(44118)/ln(3)

$${\mathtt{b}} = {\frac{{ln}{\left({\mathtt{44\,118}}\right)}}{{ln}{\left({\mathtt{3}}\right)}}} \Rightarrow {\mathtt{b}} = {\mathtt{9.734\: \!665\: \!497\: \!315\: \!512\: \!4}}$$

Because you have arithmetic mod 65537, you could find other answers by adding multiples of 65537 on to 44118 first.  For example

$${\mathtt{b2}} = {\frac{{ln}{\left({\mathtt{44\,118}}{\mathtt{\,\small\textbf+\,}}{\mathtt{65\,537}}\right)}}{{ln}{\left({\mathtt{3}}\right)}}} \Rightarrow {\mathtt{b2}} = {\mathtt{10.563\: \!412\: \!108\: \!297\: \!925\: \!2}}$$

$$\left({{\mathtt{3}}}^{\left({\frac{{ln}{\left({\mathtt{44\,118}}{\mathtt{\,\small\textbf+\,}}{\mathtt{65\,537}}\right)}}{{ln}{\left({\mathtt{3}}\right)}}}\right)}\right) {mod} \left({\mathtt{65\,537}}\right) = {\mathtt{44\,117.999\: \!999\: \!999\: \!96}}$$

(A small numerical error creeping in there!)