Problem 60 - Precalculus with Limits

Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. \(\log _{2} x+\log _{2}(x+2)=\log _{2}(x+6)\)

Step-by-Step Solution

Answer

The solution to the logarithmic equation is \( x = 2 \) when rounded to 3 decimal places.

Step 1: Combine logarithmic expressions

Using the property of logarithms that states the sum of two logs is equivalent to the log of the product of two numbers, combine the left side of equation, to form: \( \log _{2}(x(x + 2)) = \log _{2}(x + 6) \).

Step 2: Simplify logarithmic expressions

Using the property that if \( \log_b(a) = \log_b(c) \), then \( a = c \), simplify the equation to get \( x(x + 2) = x + 6 \).

Step 3: Distribute x and simplify the equation

Distribute x on the left side of the equation to get \( x^2 + 2x = x + 6 \). Move the terms involving x to the left to get \( x^2 + x - 6 = 0 \).

Step 4: Find the roots of the equation

Factor the quadratic equation to find the roots. \( (x - 2)(x + 3) = 0 \). The roots of the equation are x = 2 and x = -3.

Step 5: Eliminate extraneous solutions

In a logarithmic equation, the argument of the log function (the expression inside the log) must always be positive. Therefore, we must reject the solution x = -3, which doesn't satisfy this condition. The solution x = 2, however, does satisfy the condition, and is thus the solution to the logarithmic equation.

Key Concepts

Logarithmic Properties

Understanding logarithmic properties is essential when solving logarithmic equations. Logarithms are the inverses of exponential functions, and they come with a set of properties that make it easier to manipulate and solve logarithmic expressions. Two fundamental properties often used in algebra are the product rule and the property of equality.

The product rule allows us to combine two logarithms with the same base into a single log by multiplying their arguments. In mathematical terms, this is expressed as \( \log_b(m) + \log_b(n) = \log_b(mn) \). Another critical property is the equality property, which states that if \( \log_b(a) = \log_b(c) \), then \( a = c \). This allows you to remove the logarithmic function from both sides of the equation when the bases and the 'log' parts of the equation are the same.

In our example, these properties enable us to combine the separate logarithmic terms and then drop the log altogether, narrowing it down to a basic algebraic equation that we can solve.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, typically in the form \( ax^2 + bx + c = 0 \), where \( a \) is not equal to zero. Solving quadratic equations is a fundamental algebra skill. There are several methods to find the roots (solutions) of a quadratic equation, such as factoring, completing the square, using the quadratic formula, or graphing.

In the given problem, after applying logarithmic properties, we obtain a quadratic equation which we then solve by factoring. Factoring involves finding two binomials that when multiplied together, give the original quadratic expression and when set equal to zero, can give us the potential solutions for \( x \). These solutions are critical points where the quadratic equation intersects the \( x \)-axis on a graph.

Algebraic Manipulation

Algebraic manipulation encompasses the various techniques used to rearrange and simplify equations. It involves operations such as expanding, factoring, distributing, combining like terms, and canceling. Mastery of these techniques allows students to effectively transform complex expressions into more workable forms.

In solving our logarithmic equation, we performed algebraic manipulation by distributing \( x \), combining like terms, and moving terms across the equation to isolate the variable \( x \) on one side. Such manipulation paved the way for us to recognize the equation as a quadratic equation and then proceed to factor it appropriately.

Extraneous Solutions

Extraneous solutions are results that emerge from the process of solving an equation but do not satisfy the original equation. They commonly occur in logarithmic and radical equations because certain algebraic operations, like squaring both sides or using logarithmic properties, can introduce solutions that were not part of the original solution set.

In our case, once the equation was simplified to a quadratic equation and solved, we obtained two potential solutions. However, with logarithmic equations, we must remember that the arguments (the input of the log function) must be positive. Therefore, we have to check each solution and discard any that make the argument negative or zero. This is the critical step where we identify and eliminate extraneous solutions, ensuring our final answer is valid in the context of the original problem.