Problem 85 - Elementary and Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ e^{3 x}=9 $$

Step-by-Step Solution

Answer

Exact solution: \( x = \frac{\ln(9)}{3} \); Approximate solution: \( x \approx 0.7324 \).

Step 1: Apply the Natural Logarithm

To solve the equation \( e^{3x} = 9 \), we first apply the natural logarithm (ln) to both sides of the equation to eliminate the exponential term. This gives us \( \ln(e^{3x}) = \ln(9) \). Since \( \ln(e^y) = y \), we can simplify the left side to \( 3x \):\[ 3x = \ln(9) \]

Step 2: Isolate the Variable

Next, we need to solve for \( x \) by isolating it. We do this by dividing both sides of the equation by 3:\[ x = \frac{\ln(9)}{3} \]

Step 3: Calculate the Exact Solution

Evaluate \( \ln(9) \) using a calculator or logarithm tables:\[ \ln(9) \approx 2.1972 \]

Step 4: Obtain the Final Approximate Result

Substitute the value of \( \ln(9) \) back into the equation to find \( x \):\[ x = \frac{2.1972}{3} \approx 0.7324 \]This is the approximate solution to four decimal places.

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm to the base \( e \), where \( e \approx 2.71828 \). It is a mathematical function used to solve equations involving exponential terms. The beauty of the natural logarithm is its ability to simplify expressions with exponentials. For instance, when you take the natural logarithm of \( e \) raised to any power, it simplifies directly to that power. This property can be expressed as \( \ln(e^y) = y \).
This unique feature is incredibly useful because it allows us to solve equations containing exponentials by linearizing them. In the exercise given, \( e^{3x} = 9 \), applying the natural logarithm to both sides helps us remove the exponential, leaving us with \( 3x = \ln(9) \).
This simplification makes it much easier to proceed in finding the solution to the equation.

Isolating Variables

Isolating variables is a fundamental skill in algebra used to solve for unknowns. The goal is to have the variable of interest alone on one side of the equation, typically the left, while every other term is shifted to the other side. To achieve this, we use algebraic operations like addition, subtraction, multiplication, and division.
In our example, after applying the natural logarithm and simplifying, we have the equation \( 3x = \ln(9) \). Our next step is to isolate \( x \) by dividing both sides by 3. This gives us \( x = \frac{\ln(9)}{3} \).
By doing so, we can clearly see what \( x \) is in terms of known quantities, which is crucial before calculating any exact or approximate numerical solutions.

Exact and Approximate Solutions

When solving equations, obtaining both exact and approximate solutions can be helpful. An exact solution is expressed in a form that precisely represents the answer. In our example, \( x = \frac{\ln(9)}{3} \) is the exact solution; it is the most accurate representation involving mathematical constants or operations.
Sometimes, you might be asked to find an approximate solution, which provides a numerical estimation of the exact solution. This involves calculating values to a certain degree of precision, often useful in real-world applications where specific decimal precision is necessary. In this scenario, we compute \( \ln(9) \approx 2.1972 \). Plugging this value into our expression for \( x \) yields \( x = \frac{2.1972}{3} \), giving an approximate solution of \( x \approx 0.7324 \).
Having both solutions allows for flexibility and understanding, as exact solutions give insight into the underlying mathematical relationships, while approximate solutions provide practical values for use in calculations.