Problem 112 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the logarithmic equation algebraically. Round the result to three decimal places. Verify your answer(s) using a graphing utility. $$\ln (x+1)-\ln (x-2)=\ln x$$

Step-by-Step Solution

Answer

The only solution to the equation is \(x = 2\), while \(x = 1\) is an extraneous solution and was discarded due to the domain restrictions of the original logarithmic equation.

Step 1: Apply Logarithms Property

First, apply the property of logarithms which states that the difference of two logs is equivalent to the log of the quotient of their arguments, in this case it means transforming the equation into: \(\ln\left(\frac{x+1}{x-2}\right)=\ln x.\)

Step 2: Equate Arguments

Given that both sides are natural logs with the same base, the arguments must be equal. This gives the equation: \(\frac{x+1}{x-2}=x\).

Step 3: Simplify Equation

Solve this rational equation by multiplying both sides by \((x-2)\) to clear the fraction, then collect all terms on one side of the equation to zero (this gives a quadratic equation), and finally factor and solve. Doing this, we get: \(x^2 - 3x + 2 = 0\).

Step 4: Solve the Quadratic Equation

This quadratic equation can be factored as \((x-2)(x-1) = 0\). So the roots of the equation are \(x = 2\) and \(x = 1\).

Step 5: Check for Extraneous Solutions

The domain restrictions of the original equation, \(\ln (x+1)-\ln (x-2)=\ln x\), are \( x > 2\). Thus, \(x = 1\) is not a valid solution and \(x = 2\) stands as the only solution.

Step 6: Verify the Solution

To verify the solution, substitute \(x = 2\) into the original equation and ensure both sides are equal or graph the function \(\ln (x+1)-\ln (x-2)-\ln x = 0\) and inspect whether the function crosses the x-axis around \(x = 2\).

Key Concepts

Logarithmic Properties

Logarithmic properties are essential tools in simplifying and solving logarithmic equations. One fundamental property we used here is the difference of logarithms, which equates to the logarithm of a quotient. For example, the expression \(\ln(a) - \ln(b)\) is equivalent to \(\ln\left(\frac{a}{b}\right)\). This property allows us to combine multiple log expressions into one, making the equation easier to handle.

In the exercise, this property simplifies the original equation \(\ln(x+1) - \ln(x-2) = \ln x\) to \(\ln\left(\frac{x+1}{x-2}\right) = \ln x\). By having a single logarithm on each side, we can equate the arguments directly if the bases are the same, leading to simpler algebraic manipulation.

Natural Logarithms

Natural logarithms use the base \(e\), a special mathematical constant approximately equal to 2.718. The natural log, denoted as \(\ln(x)\), helps in exponential growth and decay problems. It is particularly useful in scientific fields because of its natural occurrence in mathematics.

In the problem, natural logarithms enable us to equate and manipulate both sides effectively. When dealing with equations that involve \(\ln\), remember that the inverse operation is the exponential function \(e^x\). This relationship is crucial for checking solutions and ensuring no steps are skipped during solving.

Quadratic Equations

Quadratic equations, which follow the standard form \(ax^2 + bx + c = 0\), arise after simplifying the problem. Here, after equating arguments and clearing fractions, we get \(x^2 - 3x + 2 = 0\). This particular quadratic can be factored easily into \((x-2)(x-1) = 0\).

Factoring quadratics allows us to find the solutions or roots. For this equation, the roots are \(x = 2\) and \(x = 1\). However, solving quadratics might require different techniques if they can't be easily factored, such as completing the square or using the quadratic formula.

Domain Restrictions

Domain restrictions are crucial when dealing with logarithmic functions. The natural logarithm \(\ln(x)\) is only defined for positive \(x\). In the equation \(\ln(x+1)-\ln(x-2)=\ln x\), we must ensure all expressions inside the logarithms are positive.

This requirement leads to the restriction \(x+1 > 0\), \(x-2 > 0\), and \(x > 0\), simplifying to \(x > 2\). After solving for \(x\), we find the solutions \(x = 2\) and \(x = 1\), but \(x = 1\) doesn't satisfy the restriction, leaving \(x = 2\) as the valid solution. Always check domain constraints to avoid extraneous solutions.

Graphing Utilities

Graphing utilities, like graphing calculators or software, offer a visual way to verify solutions. By graphing the function \(\ln(x+1)-\ln(x-2)-\ln x = 0\), we can inspect where the graph intersects the x-axis.

An intersection at the x-axis indicates a valid solution to the equation. In this case, looking around \(x = 2\) confirms our algebraic finding. These visual aids are excellent for verifying and understanding equations, especially when tackling complex problems.