Answer
By the definition of $\Theta$-notation, $f(x)$ 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 $x>0$, $\log_2x \leq x$
$5x\log_2x \leq 5x^2$ (multiply both sides by $5x$)
$x^2+5x\log_2x \leq 6x^2$ (add $x^2$ to both sides)
$|x^2+5x\log_2x| \leq 6|x^2|$ (because all terms are positive)
For $x>1$, $0k$.
Thus $x^2+5x\log_2x$ is $\Theta(x^2)$.
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, $\log_2x≤x$ because 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.
