Prove that $\frac{1}{x}-\frac{\ln 2}{2^x} > 0$ if $x>0$

Question

Given that $x>0$, can we show that

$$\dfrac{1}{x}-\dfrac{\ln 2}{2^x} > 0$$

I've plotted it out and it appears to be monotonically approaching its $0$ limit, and when plugged into Wolfram it appears to have no real $0$s, but I do not know how to go about this analytically to prove it. This is the final step in a long sequence of steps, but this inequality is fairly self-contained so I didn't think it relevant to include said extra information.

Answer 1

The given inequality is equivalent to \begin{align*} \frac{1}{x} - \frac{\ln(2)}{2^{x}} > 0 \Longleftrightarrow 2^{x} > \ln(2^{x}). \end{align*} which is true indeed. Such result comes from the fact that $y > \ln(y)$ for any $y\in(0,+\infty)$.

Answer 2

We have: $\dfrac{1}{x} - \dfrac{\ln 2}{2^x}= \dfrac{2^x-\ln(2^x)}{x\cdot 2^x}> 0$ because $x > 0, 2^x > 0$ and $2^x > \ln(2^x)$. The last one holds because $y > \ln y$ is true due to $e^y > y$ which is clearly true for $y = 2^x > 0$ .

Answer 3

Note that, for $x>0$,

$$2^x - x\ln 2=\ln2 \int_0^x (2^t-1)dt + 1 >1 > 0$$

Rearrange to obtain

$$\frac{1}{x}-\frac{\ln 2}{2^x} > 0$$