How do I show that $\log_2(k)$ is an irrational number if $k$ is an odd integer greater than $1$?

Question

Show that if $k$ is an odd integer grater than $1$, then $\log_2(k)$ is an irrational number.

So my approach for this problem was to use proof by contradiction so I wrote "Assume by contradiction if $k$ is an odd integer greater than $1$ than $\log_2(k)$ will be a rational number. Such that $\log_2(k)=a/b$, $\gcd(a,b) = 1$ and $a,b$ are integers in its lowest terms. But I don't know what to do after. Should I use the fundamental theorem of arithmetic? Euclid's lemma?

Answer 1

$2^{(a/b)} = k \implies 2^a = k^b.$

But $2^a$ is even and $k^b$ is odd.

Answer 2

Result: If $k$ is an odd integer greater than $1$, then $\log_2(k)$ is irrational.

Proof: Suppose, to the contrary, that $\log_2(k)$ is rational. In other words, $$\log_2(k)=\frac{a}{b}$$ for some $a,b\in\mathbb{Z}$ with $b\neq 0$. Exponentiating both sides of the above equation base 2, we have \begin{align*} k &= 2^{a/b} \\ k^b &= 2^a \\ &= 2\cdot 2^{a-1}. \end{align*} Since $2^{a-1}\in\mathbb{Z}$, $k^b$ must be even. However, $k$ is odd, so this is impossible.$\ \ \ \ \ \blacksquare$