Answer
Show $2x + \log_2x$ is $\Theta(x)$.
By the definition of $\Theta$, $f(x)$ is $\Theta(g(x))$ iff there exist positive real numbers $A, B, k,$ such that $A|g(x)| \leq |f(x)| \leq B|g(x)|$ for all $x>k$.
For all $x>0$, $\log_2x \leq x$ (see graph below).
Add 2x to both sides: $2x + \log_2x \leq 3x$
All terms are positive so: $|2x+ \log_2x| \leq 3|x|$.
For $x>1$, $0 < \log_2x$.
Add $2x$ to both sides: $2x < 2x+ \log_2x$.
Since all terms are positive, this is equivalent to: $2|x| < |2x+\log_2x|$.
Thus let k=1, A=2, B=3. Then for all positive real numbers $x>k,$ $A|x| \leq |2x + \log_2x| \leq B|x|$.
Hence by the definition of $\Theta,$ $f(x)$ is $\Theta(g(x))$.
Work Step by Step
The proof explains the step by step process to arrive at the derivation. Here are supplementary details to the proof:
For all $x>0$, $\log_2x \leq x$ (see graph). Also recall by statement 11.4.11: $\log_bx$ is $O(x^r)$ for all real numbers $b >1$ and $r>0$. In this case b=2, r=1.
