Problem 100 - Elementary and Intermediate Algebra

Question

Use a graphing calculator to solve each equation. If an answer is not exact, round to the nearest tenth. See Using Your Calculator: Solving Exponential Equations Graphically or Solving Logarithmic Equations Graphically. $$ \ln (2 x+5)-\ln 3=\ln (x-1) $$

Step-by-Step Solution

Answer

The solution to the equation is \( x = 8 \).

Step 1: Simplify the Equation Using Logarithmic Properties

Use the property of logarithms that states \( \ln a - \ln b = \ln \left( \frac{a}{b} \right) \). Therefore, simplify the left side of the equation: \( \ln \left( \frac{2x+5}{3} \right) = \ln (x-1) \).

Step 2: Set the Logarithmic Expressions Equal

Since the logarithms on both sides of the equation are equal, we can set their arguments equal to each other: \( \frac{2x+5}{3} = x-1 \).

Step 3: Solve the Equation Algebraically

Solve the equation \( \frac{2x+5}{3} = x-1 \) by first eliminating the fraction. Multiply both sides by 3 to get \( 2x + 5 = 3(x - 1) \). Then expand and simplify: \( 2x + 5 = 3x - 3 \). Rearrange to isolate terms with \( x \) on one side: \( 5 + 3 = 3x - 2x \). Simplify to get \( x = 8 \).

Step 4: Verify the Solution Using a Graphing Calculator

Graph the equations \( y_1 = \ln \left( \frac{2x+5}{3} \right) \) and \( y_2 = \ln (x-1) \) on a graphing calculator. Verify that they intersect at \( x = 8 \), confirming the solution is correct.

Key Concepts

Logarithmic Properties

Logarithmic properties offer essential tools for manipulating and solving equations that involve logarithms. One of these properties is the difference of logarithms, expressed as \( \ln a - \ln b = \ln \left( \frac{a}{b} \right) \). This property helps condense expressions, making equations easier to solve.

When you come across an equation with two logarithmic terms subtracting, you can rewrite it to have a single logarithm by dividing the arguments. For example, \( \ln (2x+5) - \ln 3 \) simplifies to \( \ln \left( \frac{2x+5}{3} \right) \). This simplification makes the equation cleaner and often easier to work with in subsequent steps.

Remember that properties of logarithms are not just tricks; they stem from fundamental rules of exponents. Recognizing these log properties allows you to transform complex expressions into alternates that are more manageable, paving the way for algebraic manipulations.

Solving Equations

Once you've simplified the equation using logarithmic properties, the next step is to set the expressions equal to each other if they share the same base. For the equation \( \ln \left( \frac{2x+5}{3} \right) = \ln (x-1) \), both sides have a natural logarithm, meaning the arguments themselves can be equated: \( \frac{2x+5}{3} = x-1 \).

This equality is crucial as it eliminates the logarithms, leaving you with a straightforward algebraic equation. Solving these algebraic equations involves typical algebra steps:

• Remove fractions by multiplying through by common denominators;
• Rearrange terms to isolate the variable;
• Simplify both sides where possible.

In the given exercise, multiplying both sides by 3 clears the fraction, leading to a simpler form \( 2x + 5 = 3x - 3 \). From here, arranging terms to isolate \( x \) simplifies to \( x = 8 \). This process is the backbone of solving equations: manipulating and isolating until the variable's value becomes clear.

Exponential Equations

Though this exercise directly involves logarithmic manipulation, the principles can be applied similarly when dealing with exponential equations. Exponential equations often require converting from exponential to logarithmic form to facilitate solving.

Logarithms and exponents are inverse operations. If you have an equation like \( e^x = k \), taking the natural logarithm of both sides \( \ln(e^x) = \ln(k) \) simplifies to \( x = \ln(k) \). Here, the logarithm removes the exponential term, making it more approachable for algebraic manipulation.

Similar strategies apply to graphing these equations for solutions. In the original exercise, the use of a graphing calculator confirms the algebraic solution by visually identifying where two function graphs intersect. Graphical methods are especially valuable when solutions are not neat integers or when approximate values are acceptable in the context of exponential behaviour.

Understanding how to switch forms and apply a graphing calculator to visualize results further strengthens your proficiency in solving intricate math equations.