Problem 69 - Precalculus with Limits

Question

Use a graphing utility to graph and solve the equation. Approximate the result to three decimal places. Verify your result algebraically. \(2 \ln (x+3)=3\)

Step-by-Step Solution

Answer

Approximately, the solution is \(x = 1.182\). Cross check and verify this solution graphically and algebraically.

Step 1: Solving the Equation Graphically

To solve the equation graphically, you would first enter the equation \(2 \ln (x+3) - 3 = 0\) into the graphing utility. Once the graph is plotted, identify where the graph intersects the x-axis. These are the approximate equation solutions. Result should be approximated to three decimal places using the graph.

Step 2: Transforming the Equation Algebraically

Start solving the question algebraically, by first isolating the ln-term. Thus, perform the following operations on the equation: \((2 \ln (x+3) = 3) /2\) which gives \(\ln (x+3) = 1.5\).

Step 3: Exponentiate Both Sides

Since the base of natural log (\ln) is e, exponentiating both sides of the equation using base e eliminates the \ln on the left side. So, \(e^{\ln (x+3)} = e^{1.5}\).

Step 4: Solve for x

The left side simplifies to \(x+3\), while the right side is a constant. Therefore, the equation simplifies to \(x+3 = e^{1.5}\). Solving for x gives \(x = e^{1.5}-3\).

Step 5: Evaluate the Expression

Finally, use a calculator to approximate the value of \(x = e^{1.5} - 3\) to three decimal places.

Key Concepts

Natural Logarithms

Natural logarithms are a special type of logarithm with the base \(e\), which is an irrational and transcendental number approximately equal to 2.71828. In various mathematical applications, natural logs are quite useful, especially when dealing with growth processes. When you see \(\ln(x)\), it signifies a natural logarithm. You can think of

• \(\ln(e) = 1\), because \(e^1 = e\)
• The derivative of \(\ln(x)\) gives \(\frac{1}{x}\)
• When integrating \(\ln(x)\), you end up with \(x \ln(x) - x + C\) where \(C\) is an integration constant.

The use of natural logarithms often makes complicated exponential equations easier to solve, as it transforms multiplication into addition, offering a more straightforward path in mathematical problem-solving.

Solving Equations Algebraically

Solving equations algebraically often requires simplifying expressions and isolating terms. For the equation \(2 \ln(x+3) = 3\), the first step is an algebraic manipulation to isolate the \(\ln\) term. By dividing both sides by 2, we transform it into \(\ln(x+3) = 1.5\). This simplification preps the equation for exponentiation, a common step when working with logarithms. Exponentiating

• Removes the logarithm.
• Transforms the equation into \(x+3 = e^{1.5}\).
• Allows us to solve for \(x\) by subtracting 3, giving \(x = e^{1.5} - 3\).

These algebraic techniques help to isolate the variable of interest, making them essential when dealing with logarithmic equations.

Graphing Utilities

Graphing utilities are vital tools in visualizing equations and their solutions. They allow us to see where an equation meets certain criteria, such as intersecting the x-axis, indicating solutions to an equation set equal to zero. Using a graphing utility

• Enter the equation \(2 \ln (x+3) - 3 = 0\).
• Observe where the graph intersects the x-axis.
• Find this intersection point, which represents the solution to the equation.

The ability to graphically represent equations offers invaluable insights, making it easier to approximate solutions quickly, often to a specified level of precision, such as three decimal places in this exercise.

Approximation Techniques

Approximation techniques come in handy when finding exact solutions is challenging due to complex expressions or irrational numbers. After deriving the expression \(x = e^{1.5} - 3\), it’s necessary to use a calculator to find a numerical approximation. These techniques help to

• Estimate values that are difficult to solve analytically.
• Provide clarity by condensing continuous data into manageable figures.
• Offer a pragmatic approach in applied mathematics and real-world problems.

In this case, after calculating \(e^{1.5}\) and subtracting 3, you achieve an approximation, refined to three decimal places, offering a practical and exact solution for further analysis or practical applications.