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Is it possible to proove that solution exists for this equation ? $2^x\equiv 30930818124575525495\pmod {3^{41}}$

I know $x < 3^{41}$ if solution exists. I tried brute force approach couldn't find any solution up to $x < 10^{10}$. Any help will be appreciated

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    I haven't tried but you could try proving that there exists a solution $2^x\equiv n\pmod {3^{41}}$ for all odd numbers $n$. – 2017-02-07

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Sticking to brute force methods:

According to wolfram alpha, the order of $2$ $\pmod {3^{41}}$ is $24315330918113857602$ But this is $2\times 3^{40}=\varphi(3^{41})$. Thus $2$ is a primitive root $\pmod {3^{41}}$. As your number is not divisible by $3$, we conclude that there is indeed a solution to your congruence. Granted, this approach doesn't tell us what the solution is.

Note: computing the order of $2$ isn't that hard. After all, it has to be either $2\times 3^n$ or $3^n$ for $n\in \{0,\cdots,40\}$. Even if you just check each case (not necessary), it's only $82$ cases.

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    Just learned how to compute order on modulo, thanks very much. – 2017-02-07
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    You're welcome. Must say: this is a tough problem to assign shortly after introducing the concept. – 2017-02-07
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    What about finding actual value of x ? Is there any shortcut ? – 2017-02-08
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Hint $ $ Use this theorem: if $g$ is a primitive root modulo an odd prime $p,$ then $g$ is a primitive root modulo all powers $p^k,\,$ unless $g^{p-1}\equiv 1 \pmod{p^2}$; in that case, $\,g+p\,$ is.

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    I will look into this, thanks – 2017-02-07
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    @safarov Note that if you use this result then no large calculations are required, only a few seconds of mental arithmetic.Almost surely this is what was intended by the author of the problem (assuming it was an exercise). – 2017-02-07
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    What about finding actual value of x? Is there any shortcut ? – 2017-02-08
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    @safarov Your question asks only about *existence* of a solution - which is solved easily by the above. As for computing a solution $x$, this is a difficult problem in general, know as the *discrete logarithm* problem. Follow the link in [this answer](http://math.stackexchange.com/a/32353/242) for a good introduction. – 2017-02-08