Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$

The difference is so small that I see no other way than to do the computation. Note $$e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$$ implies $$e^{13/32} > 1 + \frac{13}{32} + \frac{(13/32)^2}{2!} + \frac{(13/32)^3}{3!} + \frac{(13/32)^4}{4!} = \frac{12591963}{8388608} > \frac{3}{2}.$$

\begin{align*} \exp\left(\frac{13}{64}\right) & = \exp\left(\frac15\right)\exp\left(\frac1{320}\right) \\ & > \left(1 + \frac15 + \frac1{50} + \frac1{750}\right)\left(1 + \frac1{320}\right) \\ & = \left(1 + \frac{166}{750}\right)\left(1 + \frac1{320}\right) \\ & = \frac{458}{375}\times\frac{321}{320} = \frac{229\times107}{125\times160} \\ & = \frac{24{,}503}{20{,}000} > \frac{24{,}500}{20{,}000} = \frac{49}{40} \\ \therefore\ \exp\left(\frac{13}{32}\right) & > \left(\frac{49}{40}\right)^2 = \frac{2{,}401}{1{,}600} > \frac32. \end{align*}

We'll prove that $\ln\frac{3}{2}<\frac{13}{32},$ for which we'll prove that for any $x\geq1$ the following inequality holds. $$\ln{x}\leq(x-1)\sqrt[3]{\frac{2}{x^2+x}}.$$ Indeed, let $f(x)=(x-1)\sqrt[3]{\frac{2}{x^2+x}}-\ln{x}.$

Thus, $$f'(x)=\frac{\sqrt[3]2(x^2+4x+1)-3\sqrt[3]{x(x+1)^4}}{3\sqrt[[3]{(x^2+x)^4}}=\frac{2(x^2+4x+1)^3-27x(x+1)^4}{someting\\positive}=$$ $$=\frac{(2x^2+5x+2)(x-1)^4}{someting\\positive}\geq0,$$ which gives $$f(x)\geq f(1)=0.$$ Thus, $$\ln1.5<0.5\sqrt[3]{\frac{2}{3.75}}=\frac{1}{\sqrt[3]{15}}.$$ Id est, it's enough to prove that: $$\frac{1}{\sqrt[3]{15}}<\frac{13}{32}$$ or $$32768<32955$$ and we are done!