Prove that: e^pi > pi^e, without using a calculator

[Parth Dave]

How can you prove that:
e^pi > pi^e,
without using a calculator. (all you know is the values for e and pi.)

 

[mathwonk]

well that says that e^pi > e^(e ln(pi)), so since e^x is increasing you just have to show that pi > e ln(pi). so we just need to show that pi/ln(pi) > e.

but x/ln(x) is increasing for x >=e, and e/ln(e) = e, and pi > e, so pi/ln(pi) > e/ln(e) = e. qed.

thats cute, I'm going to try that on my calculus class.

 

[tacman]

To prove that e^pi > pi^e without using a calculator, we will use the fact that e is approximately equal to 2.71828 and pi is approximately equal to 3.14159.

First, let's rewrite the inequality as e^(3.14159) > (2.71828)^3.14159.

Next, we can take the natural logarithm of both sides of the inequality, since the natural logarithm is a monotonically increasing function and preserves inequalities.

So, we have ln(e^(3.14159)) > ln((2.71828)^3.14159).

Using the properties of logarithms, we can simplify the left side to 3.14159 and the right side to 3.14159 * ln(2.71828).

Now, we can see that the inequality reduces to 3.14159 > 3.14159 * 1, which is true.

Therefore, we can conclude that e^pi > pi^e without using a calculator.