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. 

Note that, for $x>0$,
$$2^x - x\ln 2=\ln2 \int_0^x (2^t-1)dt + 1 >1 > 0$$
Rearrange to obtain
$$\frac{1}{x}-\frac{\ln 2}{2^x} > 0$$
