which is Greater $31^{11}$ or $17^{14}$

Question

Which is Greater $31^{11}$ or $17^{14}$

My try:

Consider $$x=11 \ln(31)=11 \ln\left(32\left(1-\frac{1}{32}\right)\right)=55 \ln 2+11\ln\left(1-\frac{1}{32}\right)$$ hence

$$x=55\ln2 -11\left(\frac{1}{32}+\frac{1}{2048}+\cdots \infty\right) \lt 55 \ln 2$$

Now consider $$y=14 \ln (17)=14 \ln\left(16 \left(1+\frac{1}{16}\right)\right)$$

hence

$$y=56 \ln 2+14\ln \left(1+\frac{1}{16}\right) \gt 56 \ln 2$$

Now we have $$55 \ln 2 \lt 56 \ln 2$$ which implies

$$11 \ln(31) \lt 14 \ln (17)$$ $\implies$

$$31^{11} \lt 17^{14}$$

is there any Elementary way to solve this?

Answer 1

Yes:$$31^{11}<(2^5)^{11}=2^{55}<2^{56}=(2^4)^{14}<17^{14}.$$

Answer 2

With $x = 17$ comparing

$$ (2x-3)^{11}\; \mbox{and}\; x^{14} \; \mbox{or equivalently}\; 2x-3\;\mbox{and}\; x^{14/11} $$

their difference is minimum at a point $x_0$ and for $x = 0\Rightarrow (-3)^{11} < 0$ hence

$$ 31^{11} < 17^{14} $$

NOTE

The minimum is calculated solving

$$ -\frac{3 \left(x^{14/11}-11\right)}{11 x^2} = 0 $$

Resuming

$$ (2x-3)^{11} < x^{14}\; \forall x >0 $$