Solving $28^x \equiv 2 \pmod{43}$ 
How do we solve $28^x \equiv 2 \pmod{43}$?

I know there are not generally efficient methods for computing the discrete logarithm which are defined for an invertible $a$ modulo $q$ by $$a \equiv t^x \pmod{q}, \quad 0 \leq x \leq q-1,$$ but I was wondering if there was a way to compute it efficiently here. 

We try to make $\,2\equiv 45\equiv \color{#c00}5\cdot \color{#0a0}3^2\,$ using factors of $\,3\,$ and $\,5\,$ from small powers of $\,28.\,$ 
$\qquad\qquad  28^3\equiv 22\equiv -\color{#0a0}3\cdot 7$
$\qquad\qquad  \color{#c00}{28^5}\equiv \color{#c00}5$
$\qquad\qquad 28^7\equiv 7$  
So we have $\,\color{#c00}5\,$ and we can make  $\ {-}\color{#0a0}3\equiv 28^3/28^7\equiv \color{#0a0}{28^{-4}}$
Therefore $\ 2\equiv \color{#c00}5\cdot (\color{#0a0}{-3})^2 \equiv  \color{#c00}{28^5}(\color{#0a0}{28^{-4}})^2\equiv 28^{-3}\equiv 28^{39},\ $ by $\,28^{42}\equiv 1\,$ by Fermat.
