Problem 40 - Precalculus with Calculus Previews

Question

Solve the given logarithmic equation. $$ \ln (x+3)+\ln (x-4)-\ln x=\ln 3 $$

Step-by-Step Solution

Answer

The solution is \( x = 6 \).

Step 1: Using the Product and Quotient Rule

We start by using the product rule to combine the first two logarithms: \( \ln(a) + \ln(b) = \ln(ab) \). Then, we apply the quotient rule for logarithms to the whole left side: \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \). Therefore, the given equation becomes: \[ \ln((x+3)(x-4)) - \ln(x) = \ln\left(\frac{(x+3)(x-4)}{x}\right) = \ln(3) \]

Step 2: Exponentiating Both Sides

Since both sides of the equation are logarithms, we can exponentiate both sides to remove the logarithm. This results in the equation: \[ \frac{(x+3)(x-4)}{x} = 3 \]

Step 3: Clearing the Fraction

Multiply both sides by \( x \) to eliminate the fraction: \[ (x+3)(x-4) = 3x \]

Step 4: Expanding the Equation

Expand the left side of the equation: \[ x^2 - 4x + 3x - 12 = x^2 - x - 12 \] which simplifies to: \[ x^2 - x - 12 = 3x \]

Step 5: Moving All Terms to One Side

Subtract \( 3x \) from both sides to set the equation to zero: \[ x^2 - x - 12 - 3x = 0 \] This simplifies to: \[ x^2 - 4x - 12 = 0 \]

Step 6: Solving the Quadratic Equation

Now, we solve the quadratic equation \( x^2 - 4x - 12 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -4 \), and \( c = -12 \). Calculating, we get: \[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 + 48}}{2} \] \[ x = \frac{4 \pm \sqrt{64}}{2} = \frac{4 \pm 8}{2} \] The solutions are: \( x = 6 \) and \( x = -2 \).

Step 7: Checking the Solutions

Check which of the solutions are valid within the domain of the original equation. Since \( \ln \) is only defined for positive numbers, all terms \( x+3 \), \( x-4 \), and \( x \) must be greater than zero:- For \( x = 6 \), \( x + 3 = 9 \), \( x - 4 = 2 \), and \( x = 6 \) are all positive, so this is valid.- For \( x = -2 \), \( x \) and \( x-4 \) aren't positive, so it is invalid.Therefore, the only solution is \( x = 6 \).

Key Concepts

Product Rule

In logarithmic equations, the product rule is immensely useful. It helps to simplify expressions where two logarithms with the same base are added. The rule states:

• \( \ln(a) + \ln(b) = \ln(ab) \)

In practice, you combine the contents inside the logarithms into a single log expression. In our problem, \( \ln(x+3) + \ln(x-4) \) combines to \( \ln((x+3)(x-4)) \).
This way, solving complex logarithmic problems becomes more straightforward by reducing the number of logarithms.

Quotient Rule

The quotient rule makes logarithmic subtraction manageable. When two logs are subtracted, the rule simplifies the expression as follows:

• \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \)

By using this rule in our example, the expression on the left becomes \( \ln\left(\frac{(x+3)(x-4)}{x}\right) \).
With fewer terms, calculations become more direct, aiding in finding solutions quickly.

Quadratic Formula

The quadratic formula is a reliable method to solve equations of the form \( ax^2 + bx + c = 0 \).
It is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Used when simplifying the logarithmic equation leads to a quadratic form, as in our case \( x^2 - 4x - 12 = 0 \).
It provided two solutions, \( x = 6 \) and \( x = -2 \), but only \( x = 6 \) was valid within the problem context.
The beauty of this formula is how it guarantees finding solutions for any solvable quadratic equation.

Domain of a Function

Understanding the domain of a function is crucial, especially with logarithms. The domain specifies the input values for which the function is well-defined, particularly its ability to return real numbers.
For logarithmic functions like \( \ln(x) \), the domain is all positive real numbers, i.e., \( x > 0 \).
In the given problem, checking the domain is essential to validate our solution since all components, such as \( x+3 \), \( x-4 \), and \( x \) must be positive.

• \( x > 4 \) which makes \( x-4 > 0 \)
• \( x > 0 \) keeping \( x \) and \( x+3 > 0 \)

Thus, only \( x = 6 \) fits the domain, ensuring the integrity of the solution. Understanding domain constraints prevents invalid solutions and reinforces logical consistency.