Does anyone know how to solve this: 3^y > y^3

I am pretty sure it's y> 3 but i cant prove it.

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- May 30th 2011, 09:37 AM #1

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- May 30th 2011, 10:35 AM #2

The graph of $\displaystyle 3^y - y^3$ shows that the only interval where $\displaystyle y^3>3^y$ is between somewhere near y=2.5 and y=3. It's obvious that the two functions $\displaystyle 3^y$ and $\displaystyle y^3$ are equal when y=3. The only way I can see to show that $\displaystyle 3^y > y^3$ when y>3 is to use calculus (which I guess isn't allowed in the pre-calculus forum). To find the other point where the two functions are equal is definitely tricky – there is no elementary way to find the point where the graph crosses the axis near y=2.5.