Find an exponent $b$ such that $4^b \equiv 34\pmod{107}$

This is an instance of the difficult discrete logarithm problem, but it's small enough that it is amenable to hand computation.

Algorithmically let's use Shanks' baby giant step. By below $2$ is a primitive root $({\rm ord}\,2 = 106)$ so $\,34 \equiv 2^{\large n},\,$ and we seek $\,34\cdot 2^{\large -11j}\equiv 2^{\large k}\,$ for $\,0\le j,k < 11,\,$ by repeatedly scaling$\,34\,$ by $\,2^{\large -11}\!\equiv \color{#0a0}{50}\,$ (by Remark) till we reach some $\,2^{\large k}\equiv 1,\color{}2,4,8,16,32,64,21,\color{#c00}{42},84,\color{#90f}{61}\,$

$$34 \overset{\large \times\color{#0a0}{50}}\to 95\overset{\large\times\color{#0a0}{50}}\to \color{#c00}{42\equiv 2^{\large 8}} $$

so $\ \smash[t]{34(\overbrace{2^{\large -11}}^{\large \color{#0a0}{50}})^{\Large\color{}2}}\equiv \color{#c00}{2^{\large 8}}\overset{\large \times\, 2^{\LARGE 22}\!}\Longrightarrow 34\equiv 2^{\large 30}\!\equiv 2^{\large 2b}\!\!\!\iff$ $\! 2b\equiv 30\pmod{\!106}\!\iff\! \bbox[5px,border:1px solid #c00]{b\equiv 15\pmod{\!53}}$

Remark $ \bmod 107\!:\,\ 2^{\large 11}\! \equiv 2(\color{#90f}{61})\equiv 15\ $ so applying Gauss's algorithm

$$2^{\large -11}\equiv \dfrac{1}{15}\equiv \dfrac{7}{105}\equiv \dfrac{-100}{-2}\equiv \color{#0a0}{50}$$

To prove $\,{\rm ord}\,2 = 106,\,$ by the Order Test it suffices to show that $\,2^{\large 106/p}\!\not\equiv 1$ for all primes $\,p\mid 106,\,$ i.e. $\,2^{\large 2}\!\not\equiv 1,\, $ $2^{\large 53}\!\not\equiv 1,\,$ which is true.