Problem 31 - Algebra and Trigonometry with Analytic Geometry

Question

Solve the equation. $$ \ln x=1-\ln (x+2) $$

Step-by-Step Solution

Answer

The solution is \(x = -1 + \sqrt{1 + e}\).

Step 1: Combine Logarithmic Terms

Begin by adding \(\ln(x+2)\) to both sides of the equation to combine the logarithmic terms. This gives us:\[ \ln x + \ln (x+2) = 1 \] By properties of logarithms, \(\ln a + \ln b = \ln(ab)\). Thus, we have:\[ \ln(x(x+2)) = 1 \]

Step 2: Exponentiate Both Sides

To eliminate the natural logarithm, exponentiate both sides of the equation with base \(e\), so that:\[ e^{\ln(x(x+2))} = e^1 \]This simplifies to:\[ x(x+2) = e \] Therefore, the equation becomes a quadratic equation:\[ x^2 + 2x - e = 0 \]

Step 3: Solve the Quadratic Equation

To solve the quadratic equation \(x^2 + 2x - e = 0\), use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a = 1\), \(b = 2\), and \(c = -e\). Substitute the values into the formula:\[ x = \frac{-2 \pm \sqrt{4 + 4e}}{2} \]Simplify the equation:\[ x = \frac{-2 \pm \sqrt{4(1 + e)}}{2} \]\[ x = -1 \pm \sqrt{1 + e} \]

Step 4: Determine the Valid Solution

Given that \(x\) must be positive for \(\ln x\) to be defined, consider the positive solution:\[ x = -1 + \sqrt{1 + e} \]The other solution, \(x = -1 - \sqrt{1 + e}\), is negative and not valid.

Key Concepts

Quadratic Formula

The quadratic formula is a crucial tool for solving quadratic equations, which take the form \( ax^2 + bx + c = 0 \). When you have a quadratic equation like this, each solution is given by:

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

This formula allows you to find the values of \( x \) that make the equation true by plugging in the coefficients \( a \), \( b \), and \( c \). The term under the square root sign, \( b^2 - 4ac \), is known as the discriminant. The discriminant helps determine the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root.
• If negative, the equation has no real roots, only complex ones.

Applying this to the given quadratic equation \( x^2 + 2x - e = 0 \), we set \( a = 1 \), \( b = 2 \), and \( c = -e \). Substituting these into the quadratic formula yields the solutions for \( x \). It's a structured approach that provides precise solutions.

Properties of Logarithms

Understanding the properties of logarithms is vital for solving equations involving logarithms. These properties enable simplification and rewriting of complex logarithmic expressions:

• Product Property: \( \ln(a) + \ln(b) = \ln(ab) \)
• Quotient Property: \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \)
• Power Property: \( n\ln(a) = \ln(a^n) \)

In the original exercise, you add \( \ln(x+2) \) to both sides. By using the product property, two logarithmic terms were combined into one: \( \ln x + \ln (x+2) = \ln(x(x+2)) \). This property is especially useful here, as it converts addition of logs into a single log expression making it possible to further simplify by getting rid of the logarithm completely.

Exponentiation

Exponentiation is the inverse operation of taking logarithms, often used to "undo" a logarithm in an equation. When you have an equation in terms of logarithms like \( \ln a = b \), you can exponentiate both sides to solve for \( a \):

• \( e^{\ln a} = e^b \) implies \( a = e^b \)

This principle applies directly to the equation \( \ln(x(x+2)) = 1 \) from the exercise. By exponentiating both sides with base \( e \), the natural logarithm \( \ln \) is effectively canceled out:

• \( e^{\ln(x(x+2))} = e^1 \)
• \( x(x+2) = e \)

Exponentiation here allows us to transition from a logarithmic form to an algebraic one, where the variable \( x \) can be isolated using other algebraic techniques, such as factorization or applying the quadratic formula.