How to Find an exponent of a power in a congruence

Question

I need to solve: $2^k\equiv7(\text{mod }30)$. Also there exists a homomorphism $\varphi: U_{30}\rightarrow U_{30}$ such that $\text{Ker}(\varphi)={\{\bar{1},\bar{11}\}} $

our teacher chalenged us also to solve: $x^k\equiv7(\text{mod }30)$, i.e.: find a method to find an x that for whom there exists a k that solved this congruence.

Answer 1

Your first congruence has no solutions. Note that for any $k\in\mathbb{N}$ we get that $2^k$ is even and hence the remainder after dividing it by $30$ will always be an even number, so it will never be $7$. Try to use that idea for the second part where you have to find $x$.

Answer 2

$$2^k\equiv7(\text{mod }30)$$ $$2^k-7\equiv0(\text{mod }30)$$

it means that $$2^k-7=30m\hspace{15pt} m\in Z$$

where right side is even and left side is odd means it is not possible