> 2^x = 88 (mod 167)

One solution is x=8810. There is a smaller

> 2^x = 70 (mod 83)

>

> Is there an integer x which

> satisfies the above 2 congruences

> (simultaneously)?

>

> I feel there exists no such x,

> how do I prove it!

solution; finding it is left up to the reader. :)

Each congruence above is a discrete log problem. Solve both,

then use Chinese Remainder Theorem to find the overall solution.