Problem 47 - College Algebra

Question

For the following exercises, 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 the Logarithmic Property

Apply the logarithmic property, which states \( \ln(a) + \ln(b) = \ln(ab) \). This allows us to combine the two logarithms on the left side of the equation:\[ \ln(7) + \ln(2 - 4x^2) = \ln(14) \] becomes \[ \ln(7(2 - 4x^2)) = \ln(14) \].

Step 2: Cancel the Logarithms

Since the logarithms are equal, the expressions inside them must also be equal. Thus, you can set the content of the logarithms equal to each other:\[ 7(2 - 4x^2) = 14 \].

Step 3: Simplify the Equation

Distribute the 7 across the terms inside the parentheses and simplify:\[ 14 - 28x^2 = 14 \].

Step 4: Isolate the Quadratic Term

Subtract 14 from both sides to get:\[ -28x^2 = 0 \].

Step 5: Solve for x

Since there is only one term on the left side now, divide both sides by -28:\[ x^2 = 0 \]. Take the square root of both sides to find:\[ x = 0 \].

Key Concepts

Logarithmic Properties

Logarithmic functions help simplify complex multiplications and divisions by turning them into additions and subtractions. A crucial property you should remember is: \( \ln(a) + \ln(b) = \ln(ab) \). This is called the Product Property of logarithms and is used to combine two logarithms into one.

• This property is particularly useful in solving logarithmic equations.
• By simplifying the two logarithms on the left side of our original equation \( \ln(7) + \ln(2 - 4x^2) = \ln(14) \) into \( \ln(7(2 - 4x^2)) = \ln(14) \), we make the problem much easier to manage.
• This simplification allows us to equate the insides of both logarithms, greatly reducing the complexity of the equation.

Understanding and applying logarithmic properties effectively can make solving these equations much more straightforward.

Quadratic Equations

Quadratic equations appear when we have expressions mixing squares of variables with either constants or other linear terms. They usually take the form, \( ax^2 + bx + c = 0 \). In the exercise, after we canceled out the logarithms, we dealt with\( 7(2 - 4x^2) = 14 \), which simplifies to \( -28x^2 + 14 = 14 \).

• Once simplified to \( -28x^2 = 0 \), we have a quadratic equation with no linear term \((bx)\).
• Such equations can sometimes be reduced even further to simpler forms by factoring or using algebra.
• This specific example showcases a special case where a square equals zero, which implies that the variable must also be zero. This can sometimes prevent the need for complex solutions.

Isolating Variables

One of the central tasks in solving equations is isolating the variable to one side of the equation. This often means moving all other terms to the other side. In our example, this happened when we rewrote \( 14 - 28x^2 = 14 \) by subtracting 14 from both sides to isolate the quadratic term, resulting in \( -28x^2 = 0 \).

• This step sets the stage for solving the equation for the variable \( x \).
• Especially in equations where the variable is raised to a power or involved in complex operations, isolating the term with the variable first simplifies further solving steps.
• In our example, once \( x^2 \) was isolated, dividing both sides of the equation by -28 quickly led us to the solution \( x^2 = 0 \).
• Then, taking the square root reveals the value of \( x \), reaching the final answer: \( x = 0 \).

Having efficient strategies for isolating variables makes solving equations more manageable and less overwhelming.