Problem 60 - Algebra 2

Question

Solve each equation. Round to the nearest ten-thousandth. \(\ln x+\ln (x+4)=\ln 5\)

Step-by-Step Solution

Answer

The solution is \( x = 1 \).

Step 1: Apply the properties of logarithms

Use the property of logarithms that states \( \ln a + \ln b = \ln (ab) \) to combine the logarithmic expressions on the left-hand side of the equation. This gives you: \[ \ln(x(x+4)) = \ln 5 \].

Step 2: Exponentiate both sides

To remove the natural logarithm, exponentiate both sides of the equation. This results in: \[ x(x+4) = 5 \].

Step 3: Expand the equation

Expand the left side of the equation: \( x(x+4) = x^2 + 4x \). The equation becomes \( x^2 + 4x = 5 \).

Step 4: Rearrange into a standard quadratic equation

Subtract 5 from both sides to form a standard quadratic equation: \( x^2 + 4x - 5 = 0 \).

Step 5: Solve the quadratic equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 4 \), and \( c = -5 \). Calculate the discriminant: \( b^2 - 4ac = 4^2 - 4 \times 1 \times (-5) = 16 + 20 = 36 \).

Step 6: Calculate the roots

Calculate the roots using the quadratic formula: \( x = \frac{-4 \pm \sqrt{36}}{2 \times 1} = \frac{-4 \pm 6}{2} \). This gives two solutions: \( x = 1 \) and \( x = -5 \).

Step 7: Evaluate the solutions

Since \( x = -5 \) does not satisfy the original logarithmic expression (logarithms of negative numbers are not defined in the real number system), discard \( x = -5 \). Verify that \( x = 1 \) is valid by substituting back into the original equation, which results in a true statement.

Key Concepts

Properties of Logarithms

The properties of logarithms simplify complex equations and make them easier to solve. One of the most useful properties is that the logarithm of a product can be expressed as the sum of logarithms: \( \ln a + \ln b = \ln (ab) \).

This property helps condense expressions, turning multiple logarithmic terms into a single term. For example, in the original exercise given, combining \( \ln x \) and \( \ln (x+4) \) using this property results in \( \ln(x(x+4)) \).

This simplification is crucial as it paves the way for solving the equation more efficiently. In many logarithmic equations, these properties are the keys to unlocking a straightforward method to isolate the variable.

Quadratic Formula

Once the logarithmic part of an equation is simplified and the equation is set to equal a constant, it often forms a quadratic equation. The quadratic formula is a reliable method for solving any quadratic equation of the form \( ax^2 + bx + c = 0 \).

This formula is expressed as:

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

With this formula, you can find the solutions, or "roots," for the quadratic equation.

In our exercise, the quadratic equation formed was \( x^2 + 4x - 5 = 0 \), which was then solved using this formula. Substituting the respective values, we computed the roots of the equation.

Natural Logarithms

The natural logarithm, denoted as \( \ln \), is a logarithm to the base \( e \), where \( e \) is approximately equal to 2.71828. It is widely used in mathematics and appears frequently in calculus and complex equations involving growth and decay processes.

Natural logarithms have specific properties that help simplify logarithmic equations, such as being the inverse of the exponential function. When you exponentiate a natural logarithm, like \( e^{\ln x} \), the result is simply \( x \).

This property was useful in the original exercise because exponentiating both sides of \( \ln x(x+4) = \ln 5 \) led to \( x(x+4) = 5 \), effectively removing the logarithms and simplifying the equation.

Discriminant

The discriminant is part of the quadratic formula under the square root symbol: \( b^2 - 4ac \). It provides valuable information about the nature of the solutions of a quadratic equation.

Here is what the discriminant reveals:

• If it is positive, there are two real and distinct solutions.
• If it is zero, there is exactly one real solution (a repeated root).
• If it is negative, there are no real solutions, only complex ones.

In the exercise, the discriminant was calculated as 36, a positive number, indicating two distinct real solutions.

This helps us determine that the quadratic equation \( x^2 + 4x - 5 = 0 \) has two real roots, leading to potential solutions for the variable \( x \).

Real Number System Limitations

In the context of logarithmic equations, certain limitations within the real number system must be considered, particularly when dealing with logarithms of negative numbers.

Logarithms are only defined for positive real numbers. Therefore, when solving equations involving logarithms, it is essential to evaluate the results under these constraints.

• Negative or zero values within logarithmic expressions are invalid in the real number system.
• This limitation helps discard non-viable solutions during verification checks in the solution process.

For instance, one of the solutions from the quadratic equation \( x = -5 \) was discarded because you cannot compute a natural logarithm of a negative number.

This evaluation ensures that remaining solutions, like \( x = 1 \), are valid when checking against the original logarithmic equation.