find the largest number

Question

I have a question regarding this problem that is to find the largest number. Just by looking at the problem, I know the answer should be (d), but how can I prove that (d) is larger than (c) in a simple way?

Which of the following is the largest number:

$$\begin{align} &\textrm{(a)}\quad 3.14^3 &\quad\textrm{(b)}\quad 3^{3.14} &\quad\textrm{(c)}\quad 3.14^{3.14} &\quad\textrm{(d)}\quad (1/3)^{-4} \end{align}$$

Answer 1

$$3.14^{3.14}\lt 3.2^{3.5}=3.2^3\times 3.2^{1/2}\lt 3.2^3\times 4^{1/2}=32.768\times 2\lt 81= 3^4=(1/3)^{-4}$$

Answer 2

To prove $3^4 \ge 3.14^{3.14}\tag{1}$

Taking log on both sides

To prove $\frac{4}{3.14} \ge \frac{log(3.14)}{log(3)}$

$\frac{4}{3.14} = 1.274$

$\frac{log(3.14)}{log(3)}=log_{3}3+log_{3}(1.03666) =1+0.036667 \approx 1.03666$

$log (1+x)\approx x$ for any base

Thus 1.274> 1.03666.

Goes to prove the original statement (1)

Answer 3

mathlove's solution is good. I think the following is a bit easier:

$3.14^{3.14} = 3^3\times\left(\frac{3.14}{3}\right)^3\times3.14^{0.14}$. So we just need the product of the second and third factors to be no more than 3.

The second factor is $<1.1^3=1.331<1.5$. The third factor is (much) less than $4^{0.5}=2$. So the product of these factors is $<3$.