To compare e^π and π^e without a calculator, take the natural logarithm of both expressions to simplify the comparison: ln(e^π) = π and ln(π^e) = e·ln(π). The problem reduces to determining whether π > e·ln(π), which is equivalent to ln(π)/π > 1/e. Consider the function f(x) = ln(x)/x, which reaches its maximum at x = e (since its derivative f'(x) = (1 - ln(x))/x² is positive for x < e and negative for x > e). Since π > e, f(π) < f(e) = 1/e, meaning ln(π)/π < 1/e, so e·ln(π) < π. Therefore, ln(π^e) < ln(e^π), and since ln is increasing, π^e < e^π.
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Which Is Bigger: e^π or π^e? (No Calculator)
Added:Today, we want to answer a famous question. Which one is bigger? E to the power pi or pi to the power E? At first, this looks like a simple comparison, but actually it is not obvious. In [music] E to the power pi, the base is smaller, but the exponent is bigger. And in pi to the power E, [music] the base is bigger, but the exponent is smaller. So, there is a real competition here. Which one matters more? The bigger base or the bigger exponent? [music] And we are not going to use a calculator. We want to understand why one of them is really bigger. When powers are involved, a very natural idea is to use the natural logarithm, which we write [music] as ln.
Why? Because ln brings exponents down.
But before we use it, we should remember one important fact. [music] The natural logarithm has base E, and E is greater than one. So, ln is an increasing function. That means if we compare the ln of two positive numbers, the order of the numbers does not change. So, instead of comparing E to the power pi [music] and pi to the power E, we compare their ln values. For the first one, we get ln of E to the power [music] pi equals pi because ln and E cancel each other. For the second one, we get ln of pi to the power [music] E equals E times ln of pi.
So, the whole problem becomes this.
>> [music] >> Which one is bigger? Pi or E times ln of pi? Now, we want to prove that E times [music] ln of pi is less than pi. If we divide both sides by E pi, this is the same as saying ln of pi over pi is less than 1 over E. So, now the real question is, why is ln of pi over pi less than 1 over E? Here is the main idea. Look at this [music] function. f of x equals ln of x over x. This function measures how large ln of a number is compared to the number itself. And the key fact is, this function reaches its maximum [music] at x equals E. After x equals E, it starts decreasing. Why? Let's take its derivative. The derivative of ln of x over [music] x is 1 - ln of x over x squared. Now, [music] if x is less than e, then ln of x is less than 1, so the derivative is positive. But if x is greater than e, then ln of x is greater than 1, so the derivative is negative.
Therefore, the function increases up to e [music] and decreases after e. So, x = e is the highest point. Now, we know that pi is greater [music] than e, so pi is to the right of e, and after e, the function is decreasing. Therefore, f of pi is less than f of e. That means ln of pi over pi is less than ln of e over e, but ln of e is 1. So, we get ln of pi over pi is less than 1 over e, and this is exactly what we needed. So, e * ln of pi is less than pi. That means ln of pi [music] to the power e is less than ln of e to the power pi. And since ln is increasing, the original numbers have the same order. [music] So, finally, pi to the power e is less than e to the power pi. Therefore, e to the power pi is bigger.
>> [music] >> The key idea is this: when two powers are hard to compare directly, take ln, bring the exponents down, and look for the hidden structure. [music] In this problem, the hidden structure was the function ln of [music] x over x, which reaches its maximum at x = e.
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