# find the largest number

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}

• Actually one of this numbers has the biggest power-number and the base-number... May 16 '16 at 17:03

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

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 $<3$.

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)

• If you can calculate those logarithms you can calculate $3.14^{3.14}$ directly. May 16 '16 at 22:32
• @John Kugelman.It looks like a Q from a multiple-choice test with time being a factor, and calculators banned. May 17 '16 at 3:44
• I know, that's what I'm getting at. May 17 '16 at 12:45
• @John Kugelman, If you notice, nowhere in this solution do you have to calculate the logarithm but you have to know the principle to derive at the conclusion. This is the idea that I had may not be the best. Don't take it in the wrong way. May 17 '16 at 13:29