What residues?

We have $7x^{25}-10\equiv0\mod83$ so $7x^{25}\equiv10\mod83$

letting $n=x^{25}$ :

$7n=10+83m$ then $83m\equiv-10\mod 7$

Simplifying gives $-m\equiv4\mod7 \implies m=3 \mod7$

Now $7n=10+83(3)=259 \implies n=37$

So $x^{25}=37\mod 83$

Fermat's Little Theorem tells us $37^{82}\equiv1\mod83$ and by rules of exponents $(37^{82})^{r}37\equiv37\mod83$

We want the exponent on $37$ on the left to be divisble by $25$ so $82r+1\equiv0\mod25$ and then $7r\equiv24\mod25$ which means $7r=25s+24$ and subsequently that $25s+24\equiv0 \mod 7$ .

Now $4s\equiv-3\equiv4 \mod 7 \implies s\equiv1\mod7$

Now $7r=25(1)+24=49$ and $r=7$

This gives us $(37^{82})^{7}37\equiv37\mod83$ or $37^{575}\equiv37\mod83$

This is equivalent to $(37^{23})^{25}\equiv37\mod83$ and $x\equiv37^{23} \mod83$

$x\equiv37^{23}\equiv(37^{2})^{11}\cdot37\equiv41^{11}\cdot37\equiv21^{5}\cdot41\cdot37\equiv3^{5}\cdot7^{5}\cdot41\cdot37\equiv(-2)\cdot3\cdot7^{5}\cdot41\cdot37\equiv3\cdot7^{5}\cdot37\equiv28\cdot7^{5}\equiv15^{2}\cdot7^{2}\equiv22^{2}\equiv\boxed{69}\mod83$

This last line is probably a little inefficient but it is just a matter of simplifying the product and this is the method I used

69 [x^y] 25 [*] 7 [-] 10 [=] [MOD] 83 [=] gives 0.

Answer: 69

It is simple (albeit tedious) to show that the element 2 generates the group of units (mod 83), especially since 82 = 2*41 factors so simply. From this point we can solve Satyajit's congruence with a table of indices (i.e. logarithms).

That being said, I'd love to know if there is a slicker way to attack this problem.