Problem 89 - Algebra and Trigonometry Real Mathematics, Real People

Question

Think About It In Exercises \(89-92,\) place the correct symbol \(( < \text { or } > )\) between the two numbers. $$e^{\pi} \quad \pi^{e}$$

Step-by-Step Solution

Answer

\(\pi^{e} > e^{\pi}\)

Step 1: Define the Function

We can use the logarithmic function to help us compare these two numbers. we define a function \(f(x) = xe^{1/x}\). The two given expressions can be rewritten in terms of \(f\), such that \(e^{\pi} = f(\pi)\) and \(\pi^{e} = f(e)\).

Step 2: Determine the Maximum of the Function

To find the maximum of \(f(x)\), we will differentiate \(f(x)\). Let us denote \(f'(x)\) as the derivative of \(f(x)\), which is \(f'(x) = e^{1/x} (1 - \frac{1}{x^2})\). Setting \(f'(x) = 0\), we find that the maximal value of function \(f\) is obtained for \(x=e\).

Step 3: Compare the two quantities

As \(f(x)\) reaches its maximum value at \(x=e\), we conclude that \(\pi^{e} = f(e)\) is greater than \(e^{\pi} = f(\pi)\).

Step 4: Result

So, \(\pi^{e} > e^{\pi}\).

Key Concepts

Logarithmic Functions

Logarithmic functions are incredibly handy when it comes to comparing powers or exponents, especially when the bases and exponents are not the same. They help simplify complex exponential equations into more manageable forms. When you take the natural logarithm of numbers, you can turn an exponential equation into a linear equation. This is why logarithms can be used to compare large numbers. In this exercise, we define a function using a logarithmic approach to break down the comparison of two seemingly complex numbers: \(e^\pi\) and \(\pi^e\). Here, our function is \(f(x) = xe^{1/x}\), which serves to transform and solve this problem easily through differentiation. By representing \(e^\pi\) and \(\pi^e\) in terms of the same function, we can effectively compare them and determine which is greater.

Differentiation

Differentiation is a key concept in calculus, which helps us understand how functions change and behave. In simple terms, differentiation is a process that finds the rate at which a quantity changes. The derivative of a function, represented as \(f'(x)\), gives us this rate of change. In our exercise, the function \(f(x) = xe^{1/x}\) was differentiated to find its maximum or minimum values. The derivative we calculated was \(f'(x) = e^{1/x} \left(1 - \frac{1}{x^2}\right)\). By setting \(f'(x) = 0\), we could solve to find critical points of the function. At these critical points, the function might reach a peak (maximum) or a trough (minimum). We found that the critical point was \(x = e\), indicating where the function \(f(x)\) achieved its maximum value.

Maxima and Minima

Maxima and minima are the highest or lowest points in a function, crucial for understanding where a function peaks or sinks. To identify these, we often look at the derivative of the function. When the derivative is zero, the function may either be at a maximum or minimum point. In our step-by-step solution, after differentiating \(f(x) = xe^{1/x}\), we identified that the maximum point was at \(x = e\). By verifying with a second derivative test or checking surrounding values, we confirm this is indeed a maximum. This key step enables us to compare the values \(e^\pi\) and \(\pi^e\) effectively. Knowing that the maximum occurs at \(x = e\), and \(\pi > e\), it conclusively showed that \(\pi^e\) is greater than \(e^\pi\). Understanding maxima and minima helps solve many real-life optimization problems where we need to maximize efficiency or minimize costs.