# Determining the bigger of two numbers : $\left(\frac12\right)^e$ or $\left(\frac1e\right)^2$

The question says - use the function $f(x)=sin(x)^{sin(x)}$, where $0<x<\pi$, to determine the bigger of two numbers: $\left(\frac12\right)^e$ or $\left(\frac{1}{e}\right)^2$. Any tips on how to proceed?

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I think it another way: $f(x)=x^x$ has minimum at $x=1/e$, so $(1/e)^{1/e} < (1/2)^{1/2}$. So $(1/e)^2 < (1/2)^e$. – Hanul Jeon Feb 19 '13 at 3:42
that worked----! but $sin(x)$ doesn't matter, does it? That thing is to tell us that $x$ in the expression would lie in$[-1,1]$ – Ashish Gaurav Feb 19 '13 at 3:47

$$\left(\frac{1}{2}\right)^e>\left(\frac{1}{e}\right)^2\Longleftrightarrow e^2>2^e\stackrel{\text{apply log in both sides}}\Longleftrightarrow 2\log e>e\log2\Longleftrightarrow$$

$$\frac{1}{e}=\frac{\log e}{e}>\frac{\log 2}{2}$$

Now just check that the function

$$f(x):=\frac{\log x}{x}$$

attains its maximum at $\,x=e\,$ ...

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different yet innovative way,Thanks!! – Ashish Gaurav Feb 19 '13 at 3:54
That's what I got,which is reassuring to me. Nice job,Don! – Mathemagician1234 Feb 19 '13 at 4:04