Problem 23 - Algebra A Combined Function

Question

Solve each equation. $$ \log _{4} x+\log _{4}(x+6)=2 $$

Step-by-Step Solution

Answer

The solution is \( x = 2 \).

Step 1: Apply the Product Rule for Logarithms

Recall that the logarithm product rule states that \( \log_b(a) + \log_b(c) = \log_b(a \cdot c) \). Therefore, you can combine \( \log_{4}x + \log_{4}(x+6) \) as \( \log_{4}(x(x+6)) \).

Step 2: Use the Exponential Form of Logarithms

Set the combined log equation equal to 2, which is given as \( \log_{4}(x(x+6)) = 2 \). Convert this logarithmic equation to its exponential form: \( x(x+6) = 4^2 \).

Step 3: Solve the Quadratic Equation

Calculate \( 4^2 = 16 \), so the equation becomes \( x(x+6) = 16 \).Expand and rearrange it to form a quadratic equation: \( x^2 + 6x - 16 = 0 \).

Step 4: Factor the Quadratic Equation

Factor the quadratic equation \( x^2 + 6x - 16 = 0 \). Look for two numbers that multiply to -16 and add to 6. These numbers are 8 and -2, so the equation factors as \( (x + 8)(x - 2) = 0 \).

Step 5: Solve for the Roots

Set each factor equal to zero and solve for \( x \): - \( x + 8 = 0 \) gives \( x = -8 \).- \( x - 2 = 0 \) gives \( x = 2 \).

Step 6: Validate the Solutions

Since logarithms of negative numbers are not defined in the real number system, discard \( x = -8 \). Check \( x = 2 \) in the original equation: \( \log_4(2) + \log_4(8) = 1 + 1 = 2 \), which is correct.

Key Concepts

Logarithm Product Rule

The logarithm product rule is a helpful property when dealing with logarithmic equations. When you see expressions like \( \log_b(a) + \log_b(c) \), you can simplify them to be just one logarithm: \( \log_b(a \cdot c) \).

This rule allows you to combine multiple logarithms into a single term, which simplifies the process of solving equations. It works because logarithms can be thought of as exponents, and adding exponents means multiplying their bases.

In our example, we use the product rule to combine \( \log_4x + \log_4(x+6) \) into \( \log_4(x \cdot (x+6)) \) or \( \log_4(x^2 + 6x) \). This gives us a more straightforward equation to work with, leading us to the next step in solving it.

Exponential Form of Logarithms

Once you've simplified the logarithmic equation using the product rule, the next step is to convert it into its exponential form. This usually involves identifying what your logarithmic expression equals, then rewriting it without the logarithms.

For instance, when you have \( \log_4(x(x+6)) = 2 \), you can express it as \( 4^2 = x(x+6) \).

• Remember, if \( \log_b(y) = x \), then you can write this as \( b^x = y \).
  
• It is essentially a way of rewriting the equation so that you can solve for the variable more directly.

This transformation is crucial because exponential equations are often easier to manipulate and understand than their logarithmic counterparts, especially when trying to reach a solution involving polynomials.

Quadratic Equations

A quadratic equation is an essential concept in algebra represented by the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Quadratic equations usually have two solutions, which can include real or complex numbers.

In converting our logarithmic equation into an exponential one, we find ourselves facing \( x^2 + 6x - 16 = 0 \). This is a classic quadratic equation that we need to solve.

The solutions to these equations can be tackled using various methods, such as completing the square, the quadratic formula, or factoring. In this situation, factoring is most convenient. Solving quadratic equations is a foundational skill for many more advanced mathematical problems, making this a critical competency to develop.

Factoring Quadratics

Factoring is a technique used to solve quadratic equations by expressing them as the product of two binomials. You look for two numbers that multiply to the coefficient of the constant term but add to the coefficient of the linear term.

For the equation \( x^2 + 6x - 16 = 0 \), you need to find two numbers that multiply to \( -16 \) and add up to \( 6 \). Those numbers are \( 8 \) and \( -2 \).

• Therefore, you can write the equation as \( (x + 8)(x - 2) = 0 \).
• By setting each factor equal to zero: \( x + 8 = 0 \) and \( x - 2 = 0 \), you find the solutions \( x = -8 \) and \( x = 2 \).

It's important to always verify the solutions to ensure they are within the context of the original problem.