Problem 1 - Algebra and Trigonometry

Question

Let's solve the exponential equation \(2 e^{x}=50\). (a) First, we isolate \(e^{x}\) to get the equivalent equation ____________. (b) Next, we take ln of each side to get the equivalent equation ____________. (c) Now we use a calculator to find \(x=\) ____________.

Step-by-Step Solution

Answer

(a) \(e^x = 25\), (b) \(x = \ln(25)\), (c) \(x \approx 3.22\).

Step 1: Isolate the Exponential Term

Start with the given equation: \(2 e^{x} = 50\). To isolate \(e^x\), divide both sides by 2. This gives us \(e^x = \frac{50}{2}\). Simplify to get \(e^x = 25\).

Step 2: Take Natural Logarithm (ln) of Both Sides

To solve for \(x\), take the natural logarithm of both sides of the equation. So, we have \(\ln(e^x) = \ln(25)\). Utilize the property of logarithms: \(\ln(e^x) = x\). Thus, the equation simplifies to \(x = \ln(25)\).

Step 3: Calculate Using a Calculator

Use a calculator to find the value of \(x\). Calculate \(\ln(25)\). Rounding to two decimal places, we get \(x \approx 3.22\).

Key Concepts

Natural Logarithm

The natural logarithm, designated as ln, is a special type of logarithm that commonly appears in mathematics, particularly when dealing with exponential and logarithmic equations. It is the logarithm to the base of the mathematical constant \(e \approx 2.718\). The natural logarithm is essential because it has several unique properties that simplify the process of solving equations involving an exponential function.

In this exercise, we encountered an equation with an exponential expression \(e^x = 25\). To solve for \(x\), we used the natural logarithm on both sides of the equation. What makes the natural logarithm valuable here is its ability to transform the problem into a linear form through its connection with the base \(e\).

Some vital aspects of the natural logarithm include:

• It undoes the exponential function. Specifically, \(\ln(e^x) = x\), because when you have \(e^x\) and apply \(\ln\), you get \(x\).
• The natural logarithm is also used in calculus for differentiation and integration.

What sets the natural logarithm apart is its simplicity in handling expressions related to growth and decay, making it an indispensable tool in many fields of science and engineering.

Logarithmic Properties

Logarithms have fundamental properties that are instrumental in simplifying complex expressions, especially when solving equations. Understanding these properties can make exponential and logarithmic problems much easier to handle.

In our example, after isolating \(e^x\), we applied one of these properties to solve the equation. We converted the equation \(e^x = 25\) by taking the natural logarithm, resulting in \(\ln(e^x) = \ln(25)\). This demonstrates a key logarithmic property: the logarithm of a power simplifies by bringing the exponent down as a multiplier.

Here are some key logarithmic properties that are frequently used:

• Product Property: \(\log_b(mn) = \log_b(m) + \log_b(n)\)
• Quotient Property: \(\log_b(\frac{m}{n}) = \log_b(m) - \log_b(n)\)
• Power Property: \(\log_b(m^n) = n \log_b(m)\)
• Logarithm of One: \(\log_b(1) = 0\), regardless of the base \(b\)

Understanding these properties helps to simplify expressions making it easier to isolate variables and solve equations effectively.

Solving Equations

Solving equations is a foundational skill in algebra that involves finding the value of the unknown variable that makes the equation true. When exponential equations are involved, as in this problem, the process typically involves a series of strategic steps to simplify and solve.

The given equation was \(2e^x = 50\), an exponential equation. The first step in solving this was to isolate the term with the variable, \(e^x\), by dividing both sides by 2 to get \(e^x = 25\). This is a critical step because it reduces the equation to a form where logarithmic properties can be applied.

Next, by taking the natural logarithm of both sides, \(\ln(e^x) = \ln(25)\), and using the logarithmic property \(\ln(e^x) = x\), we simplified the equation to \(x = \ln(25)\). The use of logarithms here turns an exponential equation into a straightforward one involving an arithmetic operation.

Finally, a calculator is used to evaluate \(\ln(25)\), providing a numerical solution for \(x\). Calculating this, you get \(x \approx 3.22\).

This step-by-step problem-solving approach is powerful in handling exponential equations, where isolating the variable and using logarithmic functions reduces and simplifies the equation to a solvable form.