Problem 24 - Algebra A Combined Function

Question

Solve each equation. $$ \log _{3} x+\log _{3}(x+6)=3 $$

Step-by-Step Solution

Answer

The solution is \( x = 3 \).

Step 1: Use the Product Property of Logarithms

The product property of logarithms states that \( \log_b(m) + \log_b(n) = \log_b(m \cdot n) \). Therefore, the expression \( \log_{3} x + \log_{3} (x+6) \) can be rewritten as \( \log_{3} (x(x+6)) \).

Step 2: Set the Equation to Logarithmic Form

Replace the logarithm equation with its exponential form. You have \( \log_{3}(x(x+6)) = 3 \). By the definition of logarithms, this equation implies that the argument of the log is equal to the base raised to the power on the other side, giving us \( x(x + 6) = 3^3 \).

Step 3: Solve the Quadratic Equation

Simplify the equation from Step 2. You know \( x(x + 6) = 27 \). This simplifies to the quadratic equation \( x^2 + 6x = 27 \). Rearrange terms to form a standard quadratic equation: \( x^2 + 6x - 27 = 0 \).

Step 4: Factor the Quadratic Equation

To solve \( x^2 + 6x - 27 = 0 \), you can use factoring. Set up two binomials: \((x + 9)(x - 3) = 0\). Solve for \(x\) by setting each factor to zero: \(x + 9 = 0\) or \(x - 3 = 0\). This gives solutions \(x = -9\) and \(x = 3\).

Step 5: Check the Solutions

Logarithms are only defined for positive numbers, so discard any negative solutions from \(x = -9\). This leaves us with \(x = 3\) as the solution. Substitute \(x = 3\) back into the original equation to verify: \( \log_{3}(3) + \log_{3}(3+6) = \log_{3}(3) + \log_{3}(9) = 1 + 2 = 3 \), confirming that it satisfies the given equation.

Key Concepts

Product Property of Logarithms

The product property of logarithms is a helpful tool for simplifying log expressions. It allows us to combine two logarithms with the same base into one by turning the sum into the logarithm of a product. Specifically, if you have two logarithms added together, such as \( \log_b(m) + \log_b(n) \), you can combine them into a single logarithm: \( \log_b(m \cdot n) \).
This property can greatly simplify complex logarithmic expressions and equations.

• It is important that both logarithms have the same base.
• This property can be used only when the base and the arguments of both logs are defined (i.e., positive real numbers not equal to one).

In our example, we had \( \log_{3} x \) and \( \log_{3} (x+6) \).
Applying the product property, they combine to give us \( \log_{3} (x(x+6)) \).
This simplification is crucial as it allows us to move towards solving the equation more easily.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, recognizable by the form \( ax^2 + bx + c = 0 \).
Solving a quadratic equation typically involves finding two values of \( x \) that satisfy the equation.
Methods for solving quadratic equations include factoring, using the quadratic formula, or completing the square.

In our exercise, after applying the product property, we derived \( x^2 + 6x = 27 \).
Rearranging this gives us a standard quadratic equation: \( x^2 + 6x - 27 = 0 \).

• First, try factoring, which is often the simplest method if applicable.
• In this case, the quadratic factors to \((x + 9)(x - 3) = 0\).

Once factored, set each factor equal to zero to find potential solutions for \( x \).
This equation gives us two potential solutions: \( x = -9 \) or \( x = 3 \).
However, not all potential solutions will be valid in the context of logarithms.

Exponential Form

Exponential form is a fundamental concept when dealing with logarithmic equations. It helps in transforming logarithmic equations to a more workable structure.
The core principle here is understanding the relationship between logarithms and exponents: \( \log_b(a) = c \) translates to \( a = b^c \).
This conversion is crucial for solving equations that initially seem daunting.

• In our exercise, we moved from \( \log_{3}(x(x+6)) = 3 \) to its exponential form \( x(x + 6) = 3^3 \).
• This form is easier to manipulate algebraically, allowing us to solve for \( x \) explicitly.

By converting to exponential form, you transform the logarithmic equation into a polynomial or arithmetic equation that can sometimes be more straightforward to tackle.
In our case, \( 3^3 = 27 \), which set the stage for identifying and solving the quadratic equation.
This approach streamlines the solving process, making it accessible and manageable.