# How to prove there exists $x_0 > 0$ such that for any $x > x_0$, $ax+b\gt \log_2 x$?

Suppose if $a \gt 0$ and real $b$ are given.

How to prove that there is $x_0 \gt 0$ such that for any $x \gt x_0$ implies $ax+b-\log_2x \gt 0$?

It's so obvious but I need a hint how to do it. Maybe Taylor expansion? or derivative and see that first derivative of $f(x)=ax+b-\log_2x$ , $f'(x)\gt0$ ?

Do I need that $\lim\limits_{x\to0} f(x) = -\infty$ and $\lim\limits_{x\to + \infty}f(x) = +\infty$? If $\lim\limits_{x\to +\infty}f(x)=-\infty$ then of course $2^{Cx} \gt D \cdot x^k$ for some $x > x_0 \in \mathbb R$.

• Prove that: If $a > 0$ then there exists $x_0$ such that for any $x>x_0$: $a \cdot x + b > log_2 x$ – M_T Oct 5 '15 at 20:28

$$f'(x)=a-\frac{1}{\log 2\log x}.$$
Hint: Use de l'Hopital on $$\lim_{x\to\infty} \frac{ax + b}{\log_{2}(x)}$$