Problem 354 - Precalculus

Question

Solve the equation for \(x,\) if there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution. \(\ln (x)-\ln (x+3)=\ln (6)\)

Step-by-Step Solution

Answer

The equation has no solution within the allowable domain for the logarithms.

Step 1: Apply Logarithm Quotient Rule

Start by using the logarithm quotient rule on the left side of the equation \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \). This simplifies\[ \ln (x) - \ln (x+3) = \ln \left(\frac{x}{x+3}\right) = \ln(6). \]

Step 2: Exponentiate Both Sides

To eliminate the logarithms, exponentiate both sides of the equation,\[ \frac{x}{x+3} = 6. \]This results from the property \( \ln(a) = \ln(b) \Rightarrow a = b. \)

Step 3: Solve the Rational Equation

Multiply both sides by \(x + 3\) to solve for \(x\):\[ x = 6(x + 3). \]isolate \(x\):\[ x = 6x + 18 \] \[ x - 6x = 18 \] \[ -5x = 18 \] \[ x = -\frac{18}{5}. \]

Step 4: Consider the Solution Domain

The initial logarithm functions remain valid only when their arguments are positive. Thus, for \( \ln(x) \) and \( \ln(x+3) \), \(x > 0\). Therefore, \(x = -\frac{18}{5}\) is not in the domain of the function.

Step 5: Graphing Both Functions

Graph the functions \(y = \ln(x) - \ln(x+3)\) and \(y = \ln(6)\) using graphing paper or a graphing calculator. Observing these graphs confirms there is no point of intersection in the feasible region \(x > 0\), validating that no solution exists.

Key Concepts

Logarithm Quotient Rule

The Logarithm Quotient Rule is a fundamental property of logarithms used to simplify the expression of the difference between two logarithms. Specifically, it states that \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \). In this exercise, it is used to combine the two logarithmic terms on the left-hand side of the equation:

• \( \ln(x) - \ln(x+3) \) is simplified to \( \ln\left(\frac{x}{x+3}\right) \).

This rule is particularly useful because it helps in converting subtraction of logarithms into a single logarithm, making the equation easier to solve by applying further algebraic operations. For students, understanding this rule is crucial because it forms the basis for simplifying and solving logarithmic equations efficiently.

Exponentiation

Exponentiation is the process of raising a number to a power to eliminate logarithms from equations. In the context of this exercise, once the logarithms have been combined into a single term using the Logarithm Quotient Rule, the problem becomes managing an equation like \( \ln(a) = \ln(b) \).

• Here, you can exponentiate both sides to remove the logarithms, resulting in the simpler equation \( \frac{x}{x+3} = 6 \).

This arises from the logarithmic identity that states if \( \ln(a) = \ln(b) \), then \( a = b \). By applying exponentiation, you transition from a logarithmic function to a rational one, which is often easier to manipulate and solve. This step is crucial in easing the path toward finding the solution, or verifying the absence thereof.

Rational Equations

Rational equations are those equations that involve fractional algebraic expressions, especially those where variables appear in the denominator. In this exercise, after applying exponentiation, we deal with a rational equation \( \frac{x}{x+3} = 6 \).

• To solve, multiply both sides by \( x+3 \): this clears the fraction, leading directly to \( x = 6(x+3) \).

Further simplification involves distributing and rearranging terms to isolate \( x \). Thus, we arrive at the equation \( -5x = 18 \) and solve it by dividing, yielding \( x = -\frac{18}{5} \). Understanding rational equations is vital because they frequently appear, and proficiency in solving them is essential for tackling more complex algebraic expressions.

Domain of Logarithmic Functions

The domain of a logarithmic function refers to the set of input values for which the function is defined. For any logarithm \( \ln(x) \), the domain requires that \( x > 0 \) because the logarithm of a non-positive number is undefined in real numbers.

• In our exercise, because we are working with \( \ln(x) \) and \( \ln(x+3) \), both expressions necessitate that \( x > 0 \) and \( x+3 > 0 \).

This consideration leads to the realization that \( x = -\frac{18}{5} \) is not within the allowed domain since it is negative. Hence, verifying the solution underlines the importance of considering domain constraints whenever dealing with logarithmic functions. Ignoring this can lead to accepting invalid or non-applicable solutions, which wouldn't hold up when graphing or further verifying.