= Solution
First simplify the specified residue by <modular exponentiation>. Modulo the <prime number> $113$,
$$
102^2\equiv8,\quad102^4\equiv64,\quad102^8\equiv28,\quad
102^{16}\equiv106,\quad102^{32}\equiv49,\quad102^{64}\equiv28.
$$
Therefore $102^{70}\equiv28\cdot64\cdot8\equiv98$, so the required equation becomes $x^{37}\equiv97\pmod{113}$. Its right side is nonzero, so any solution is a <unit> modulo $113$.
By <Fermat's little theorem>, exponents of such <units> can be reduced modulo $112$. Since
$$
37\cdot109=1+36\cdot112,
$$
raising to the $109$th power is the inverse of raising to the $37$th power. There is therefore exactly one residue class of solutions:
$$
x\equiv97^{109}\equiv97^{-3}\pmod{113}.
$$
Now $97\equiv-16$ and $97^3\equiv-4096\equiv85$. Since $4\cdot85\equiv1$, we obtain
$$
\boxed{x\equiv4\pmod{113}.}
$$
As a direct check, $4^4\equiv30$, $4^8\equiv109$, $4^{16}\equiv16$, and $4^{32}\equiv30$, whence $4^{37}\equiv30\cdot30\cdot4\equiv97$.
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