How can i solve this?

[seephyro]

How can i solve, detail, this problem?
Who is bigger, 3^π or π^3? Show a solution using derivative

 

[HallsofIvy]

How can i solve, detail, this problem?
Who is bigger, 3^π or π^3? Show a solution using derivative

Well,an obvious way to do this would be to calculate those numbers! But since the problem say s "using the derivative", I would look at the function \(\displaystyle f(x)= 3^x- x^3\). When x=1, f(1)= 3^1- 1^3=2.. If you can show that the derivative is always positive, that would show that \(\displaystyle a^x\) is greater that \(\displaystyle x^a\) for x> 1.

 

[lookagain]

But since the problem says "using the derivative", I would look at the function \(\displaystyle f(x)= 3^x - x^3\).

When x=1, f(1)= 3^1- 1^3=2. If you can show that the derivative is always positive,

that would show that \(\displaystyle a^x\) is greater than \(\displaystyle x^a\) for x > 1.

It turns out the derivative, \(\displaystyle \ f'(x) \ = \ 3^x(ln(3)) \ - \ 3x^2, \ \ \) is negative at x = 2, just to mention an integer value.

At x = 3, the derivative is positive. Maybe someone can show that the derivative is always positive for x > 3,

so that \(\displaystyle a^x\) is shown to be greater than \(\displaystyle x^a\) for all real values of x > 3.