Congruence Relation with exponents and variables

Question

I am currently trying to solve a congruence relation with a constant and a variable, both of which have attached exponents. The relation is as follows:

$7^{95}\equiv x^{3} (mod 10)$

How does one go about solving for x when it has an attached exponent?

Answer 1

HINT: $3^3\equiv 7\pmod{10}$, so $7^{95}\equiv 3^{3\cdot95}\pmod{10}$.

Or you could observe that $7^2\equiv-1\pmod{10}$, so $7^4\equiv1\pmod{10}$, and easily reduce $7^{95}$ to its residue mod $10$, and then if necessary solve for the cube root by trial and error: there are only ten numbers to try, after all!

Answer 2

Hint $\rm\,\ 2,5\nmid x,\:$ so $\rm\: mod\ 10\!:\ x^4\equiv 1\:\Rightarrow\: \color{#c00}{x^{-1}}\!\equiv x^3\!\equiv 7^{95}\!\equiv 7^{\large\, -1+4\,(24)}\!\equiv \color{#c00}{7^{-1}}\Rightarrow\:x\equiv 7$

Answer 3

$7^2\equiv-1\pmod{10}, 7^4\equiv1$

So, $7$ is a primitive root $\pmod{10}$

Taking discrete logarithm on $7^{95}\equiv x^3\pmod {10}$ wit base $7$

$$95\cdot ind_77\equiv 3\cdot ind_7x\pmod 4\text{ as }\phi(10)=4$$ $$\implies 3\equiv 3\cdot ind_7x\pmod 4$$ $$\implies ind_7x\equiv1\pmod 4\text{ as }(3,4)=1$$ $$\implies x\equiv7^1\pmod{10}$$