Problem 337 - Precalculus

Question

Solve each equation for \(x.\) \(\ln (7)+\ln \left(2-4 x^{2}\right)=\ln (14)\)

Step-by-Step Solution

Answer

The solution is \(x = 0\).

Step 1: Use Logarithm Properties

Use the property of logarithms that states that the sum of logs is the log of the product: \[\ln(a) + \ln(b) = \ln(a \times b) \].Applying this to the given equation, we get: \\(\ln(7 \times (2 - 4x^2)) = \ln(14) \).

Step 2: Equate the Arguments of the Logarithms

Since the natural logarithms on both sides of the equation must be equal, their arguments must also be equal. Thus, we have: \\(7 \times (2 - 4x^2) = 14 \).

Step 3: Expand and Simplify the Equation

Expand and simplify the equation from Step 2: \\(7 \times 2 - 7 \times 4x^2 = 14 \), \\(14 - 28x^2 = 14 \).

Step 4: Solve for x

Subtract 14 from both sides of the equation: \\(14 - 28x^2 - 14 = 0 \), \\(-28x^2 = 0 \). \Now, divide by -28: \\(x^2 = 0 \). \Taking the square root of both sides, we find \(x = 0\).

Key Concepts

Properties of Logarithms

When solving logarithmic equations, understanding the properties of logarithms is crucial. These properties allow you to simplify and manipulate logarithms in equation-solving processes. One essential property is the **product property**, which states that the sum of logarithms is equal to the logarithm of the product of the numbers. Mathematically, this is expressed as \( \ln(a) + \ln(b) = \ln(a \times b) \). This property is helpful when you need to combine logarithmic terms into a single logarithm.
Another key property is the **quotient property**, which deals with the subtraction of logs of values. It indicates that the difference between two logarithms is equivalent to the logarithm of the quotient of those values: \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \).
Finally, the **power property** allows you to take a power outside a logarithm, by expressing it as a multiplication inside, like so: \( \ln(a^b) = b \times \ln(a) \).

• The product property simplifies equations by combining expressions.
• The quotient property helps in separating complex terms.
• The power property is useful for managing exponents within logarithms.

By using these properties, you can simplify complex logarithmic expressions, making it easier to equate both sides of an equation as seen in the exercise.

Natural Logarithm

The natural logarithm, often represented as \( \ln \), is a specific type of logarithm that uses Euler's number, \( e \), as its base, where \( e \approx 2.71828 \). This logarithm is fundamental in many areas of mathematics, particularly when dealing with continuous growth models.
In our problem, the natural logarithm is used to compare the log expressions, such as \( \ln(7) \) and \( \ln(14) \). The function \( \ln(x) \) calculates the power to which the base \( e \) must be raised to obtain \( x \).
Why is the natural logarithm so special?

• It naturally appears in many mathematical contexts, such as calculus and complex numbers.
• It is deeply connected to exponential growth and decay processes, making it essential in natural sciences and economics.
• Its derivative is simple, allowing for easier calculations in calculus: \( \frac{d}{dx} \ln(x) = \frac{1}{x} \).

Using the natural logarithm properties allows for straightforward manipulation of equations, serving as an efficient tool for comparison, as demonstrated when both sides of the equation were normalized using the natural logarithm.

Quadratic Equations

Quadratic equations arise when the problem boils down to a second-degree polynomial equation of the form \( ax^2 + bx + c = 0 \). In our exercise, simplifying the logarithmic equation step-by-step resulted in a quadratic-like expression, focusing on terms like \(-28x^2 = 0\).
One way to solve quadratic equations is by factoring, provided they can be easily expressed as a product of binomials. In this case, our equation after simplification \( x^2 = 0 \) directly reveals the value of \( x \).
If the equation had been more complicated, you might have needed to use:

• The **Quadratic Formula**: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Square roots to solve equations like \( x^2 = k \) for both positive and negative roots.

By taking the square root of both sides, you isolate \( x \), leading directly to the solution. This method is straightforward in this context, as it shows that \( x = 0 \) is the only solution, neatly tying the whole exercise process together.