Problem 99 - College Algebra Essentials

Question

Solve each equation. $$\ln (2 x+1)+\ln (x-3)-2 \ln x=0$$

Step-by-Step Solution

Answer

The solution for the given equation is \( x = \frac{3 + \sqrt{5}}{2} \).

Step 1: Simplify the equation using the properties of logarithms

There are few basic properties of logarithms that can be employed here: The first property is \( \ln a + \ln b = \ln (ab) \). Using this property, the first two terms of the equation can be combined as: \( \ln (2 x+1)+\ln (x-3) = \ln[(2x+1)(x-3)] \) The second property is \( n \cdot \ln a = \ln( a^n ) \). Using this property, the last term can be rewritten as: \( -2 \ln x = \ln (x^{-2}) \) Now, replace these transformed terms back in the original equation, we get: \( \ln[(2x+1)(x-3)] + \ln (x^{-2}) = 0 \)

Step 2: Combine the logarithms into a single expression

Again use the first property of logarithm i.e., \( \ln a + \ln b = \ln (ab) \) to combine the obtained equation: \( \ln[(2x+1)(x-3)(x^{-2})] = 0 \)

Step 3: Convert the log equation into an exponential equation

We know that \( \ln a = 0 \) if and only if \( a = 1 \). So let's set the expression inside the logarithm equal to one: \( (2x+1)(x-3)(x^{-2}) = 1 \)

Step 4: Solve the resulting equation

Solving this equation leads us to the final step. Expanding and rearranging gives: \( 2x (x^{-2}) - 3(x^{-2}) + x^{-1} - 1 = 0 \) Simplifying this gives us: \( 2 - \frac{3}{x} + \frac{1}{x^2} - 1 = 0 \) Multiply all terms by \( x^2 \) to get rid of denominators: \( 2x^2 - 3x + 1 - x^2 = 0 \), which simplifies to: \( x^2 - 3x + 1 = 0 \) This is a quadratic equation, which can be solved using the quadratic formula: \( x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 * 1 * 1}}{2 * 1} \) thus giving us the two solutions: \( x = \frac{3 \pm \sqrt{5}}{2} \)

Step 5: Reject extraneous solutions

Sometimes solving these equations yield solutions that are not valid in the original equation, because they might lead to the log of a negative number or zero. Checking the obtained solutions:\( x = \frac{3 - \sqrt{5}}{2} \) eventually gives us a log of negative number when plugging back into the original equation, so this solution is discarded.\( x = \frac{3 + \sqrt{5}}{2} \) is valid and does not make the arguments of any log negative, and so it is the single solution for this equation.