Show that $4x + \log_2 x$ is $\Theta(x)$

Question

Show that $4x+\log_2 x$ is $\Theta$(x) Unsure how to do this I think I've found $\Omega$(x).

Here's what I did:

$x\le4x + \log_2 x$. Let $A = 1$ and $a = 1$ $A|x| \le |4x + \log_2 x |$ x > a Hence by $\Omega$ notation $4x + \log_2 x$ is $\Omega$(x).

Answer 1

Your proof for $\Omega(x)$ is right. In order to complete it to $\Theta(x)$ consider the inequality $\ln(x)\leq x-1$ (see Prove that $\ln x \leq x - 1$) Now $\log_2(x)=\ln(x)/\ln 2$, therefore for $x\geq 1$ $$0\leq |4x+\log_2 x|=4x+\log_2 x=4x+\frac{\ln x}{\ln 2}\leq 4x+\frac{x-1}{\ln 2}\leq \left(4+\frac{1}{\ln 2}\right)x\leq 6 x.$$