solve $x^{15} \equiv 2\pmod {47}$ solve $x^{15} \equiv 2\pmod {47}$
on the solution set:
suppose that $\overline{x^{15}} = \overline{2}$ then $\overline{x^{15m}} = \overline{2^m}$
solve $15m \equiv 1 \pmod {46}\;$, gaining a solution of $m = 43$ so $15m = 46k + 1$
so $\overline{x^{15m}} = \overline{x^{46k+1}} = \overline{x} = \overline{2^{43}} = 6$
my query is that isn't $15m = 46k +1$ only in modulus 46? (as thats what we solved), how can we just plug $\overline{x^{46k+1}}$ in when we're working in modulus 47?
 A: Perhaps a little simpler in this case:
Working modulo $\;47\;$ all through:
$$x^{15}=2\implies x^{45}=8$$
But Fermat's Little Theorem gives $\;x^{46}=1\;$ , so
$$x^{45}=x^{-1}x^{46}=x^{-1}\implies x^{-1}=8\implies x=8^{-1}=6$$
A: Hint $\ 15m = 1\!+\!46k\,\Rightarrow\ {\rm mod}\ 47\!:\ x^{15m} \equiv x^{1+46k}\equiv x(\color{#c00}{x^{46}})^k \equiv x(\color{#c00}1)^k\equiv x\ $ since $\,\color{#c00}{x^{46}\equiv 1}\,$ follows by Fermat's Little Theorem (note $\,x\not\equiv 0\,$ by $\,x^{15}\equiv 2).$  
Conceptually: raise $\,x^{15} \equiv 2\pmod{47}$ to power $\,\dfrac{1}{15} \equiv m\pmod{46}\,$ to get $\, x \equiv 2^m\pmod{47}$
Remark $\ {\rm mod}\ 46\!:\ m\equiv \dfrac{1}{15}\equiv \dfrac{-45}{15}\equiv \color{#c0f}{-3}\equiv \color{#0a0}{43}.\ $ But $\,2^{\large\color{#c0d}{-3}}\equiv \dfrac{1}8\equiv\dfrac{48}8\equiv 6\pmod{47}$ computes easier than $\,2^{\large \color{#0a0}{43}}.$ This explains the algebra in Don Antonio's answer, i.e. it is equivalent to using this general method but with the least magnitude rep $\,1/15\equiv \color{#c0f}{-3}\,$ vs. $\,\color{#0a0}{43}\pmod{46}.\,$ Using least magnitude (a.k.a. balanced) reps is a standard optimization method in modular arithmetic.
A: My way isn't really different but we can start with Fermat's Little Theorem , following the hint by lab bhattacharjee.
$$x^{46} \ \equiv 1 \ (mod \ 47) $$
$$(x^{15})^3 \cdot x \ \equiv \ 1 \ (mod \ 47) $$
$$ 8x \ \equiv \ 1 \ (mod \ 47) $$
Now it's obvious x = 6 
