eπ is greater than πe because the exponential function grows faster than a power function, especially when the base is e. Let's delve into why:
Understanding the Concept
The core idea revolves around comparing the growth rates of ex and xe. To do this effectively, we can analyze a related function.
A Mathematical Approach
Consider the function f(x) = ln(x) / x. If we can show that f(e) > f(π), then it follows that ln(e) / e > ln(π) / π, which can be rearranged to π ln(e) > e ln(π), and further simplified to ln(eπ) > ln(πe). Since the natural logarithm (ln) is a monotonically increasing function, this implies eπ > πe.
Analyzing f(x) = ln(x) / x
Let's find the derivative of f(x) to understand its increasing and decreasing behavior:
- f'(x) = (1 - ln(x)) / x2
Setting f'(x) = 0, we find that 1 - ln(x) = 0, which gives us x = e.
- For x < e, ln(x) < 1, so f'(x) > 0, meaning f(x) is increasing.
- For x > e, ln(x) > 1, so f'(x) < 0, meaning f(x) is decreasing.
Therefore, f(x) reaches its maximum at x = e. Since π > e, f(π) < f(e). This confirms that ln(π) / π < ln(e) / e, and consequently, eπ > πe.
Why Does This Matter?
This example highlights a crucial principle in calculus: exponential functions with bases greater than 1 eventually outpace polynomial or power functions as x grows larger. The number e provides a specific and compelling case.
Numerical Approximation
- eπ ≈ 23.1407
- πe ≈ 22.4592
This numerical result visibly confirms that eπ > πe.
Conclusion
Because the function f(x) = ln(x)/x is decreasing for x > e, and since π > e, it follows that f(π) < f(e), which ultimately proves that eπ > πe. This demonstrates the faster growth rate of the exponential function compared to the power function.
