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

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 $y = 2^x > 0$ .
