Let $f:\mathbb{N} \rightarrow \mathbb{R^{\geq 0}}$. Let $a \in \mathbb{R^{+}}$.
Assume $\exists x_{1}, k_{0} \in \mathbb{R^{+}}, \forall n\in \mathbb{N}, n \geq k_{0} \implies f(n) \leq x_{1}n^{a}$. Prove that $\exists x_{2}, k_{1} \in \mathbb{R^{+}}, \forall n\in \mathbb{N}, n \geq k_{0} \implies \log_{2}(f(n) )\leq x_{2}\log_{2}(n)$.
I started with $$f(n) \leq x_{1}n^{a} \\ \log_{2}(f(n)) \leq \log_{2}(x_{1}n^{a}) \\ \log_{2}(f(n)) \leq \log_{2}(x_{1}) +a\log_{2}n \\$$ And then to continue with a choice of $x_{2}$, proceeded with following equation by isolating $x_{2}$. $$x_{2} = \frac{\log_{2}(x_{1}) + a\log_{2}n}{\log_{2}n} \\ x_2 = \log_{n}x_{1}+a $$ Is this choice of $x_{2}$ good enough to prove this statement?
Your $x_2$ depends on $n$, but the introduction of $\exists x_2$ comes before the introduction of $\forall n$. This is not allowed. You need to find an $x_2$ independent of $n$.