Problem 56 - Precalculus

Question

Solve the logarithmic equations exactly. $$\log _{2}(x+1)+\log _{2}(x+5)-\log _{2}(2 x+5)=2$$

Step-by-Step Solution

Answer

The solution is \( x = 5 \).

Step 1: Utilize Logarithmic Properties

Begin by using the property of logarithms that states \[ \log_b(m) + \log_b(n) = \log_b(m \cdot n) \] to combine the first two terms. Thus, we have: \[ \log_2((x+1)(x+5)) - \log_2(2x+5) = 2 \]

Step 2: Apply Logarithmic Subtraction Property

Now, apply the subtraction property of logarithms: \[ \log_b(m) - \log_b(n) = \log_b\left(\frac{m}{n}\right) \] This gives us: \[ \log_2\left( \frac{(x+1)(x+5)}{2x+5} \right) = 2 \]

Step 3: Convert Logarithmic Equation to Exponential Equation

The next step is to convert the logarithmic equation to an exponential form. Since \( 2^y = x \) is equivalent to \( \log_2(x) = y \), we have: \[ \frac{(x+1)(x+5)}{2x+5} = 2^2 \] which simplifies to: \[ \frac{(x+1)(x+5)}{2x+5} = 4 \]

Step 4: Clear the Fraction

Multiply both sides by \( 2x+5 \) to eliminate the fraction: \[ (x+1)(x+5) = 4(2x+5) \] Expand both sides to simplify: \[ x^2 + 5x + x + 5 = 8x + 20 \] Which simplifies to: \[ x^2 + 6x + 5 = 8x + 20 \]

Step 5: Form and Solve the Quadratic Equation

Rearrange the equation to form a quadratic equation: \[ x^2 + 6x + 5 - 8x - 20 = 0 \] This simplifies to: \[ x^2 - 2x - 15 = 0 \] This quadratic equation can be factored: \[ (x - 5)(x + 3) = 0 \] Thus, the solutions for \( x \) are: \[ x = 5 \quad \text{and} \quad x = -3 \]

Step 6: Validate Solutions in Context

Verify each potential solution in the context of the original logarithmic equation. For \( x = 5 \): \[ \log_2(6) + \log_2(10) - \log_2(15) = \log_2\left(\frac{60}{15}\right) = \log_2(4) = 2 \] For \( x = -3 \), it causes a negative argument in \( \log_2(x+1) \), which is not allowed. Thus, \( x = 5 \) is the only valid solution.

Key Concepts

Logarithmic Properties

In dealing with logarithmic equations, one must understand the fundamental properties of logarithms. These properties allow us to transform and simplify equations for easier solving.

One important property is the **product property** of logarithms. It states that:

• \( \log_b(m) + \log_b(n) = \log_b(m \cdot n) \)

This property helps us combine two separate logarithmic terms into a single term, simplifying complex equations.

Another important property is the **difference property**, which is used to handle subtraction between logarithms. This is expressed as:

• \( \log_b(m) - \log_b(n) = \log_b\left(\frac{m}{n}\right) \)

This property consolidates terms by creating a division inside the logarithm, facilitating a clearer path to solution. These tools are essential for successfully navigating through the manipulations required in complex logarithmic calculations.

Exponential Equations

Converting logarithmic equations to exponential form can make a difficult problem much simpler. Understanding this conversion is key to solving logarithmic equations.

The concept relies on the fundamental relationship between logarithms and exponents, which is:

• If \( \log_b(x) = y \), then \( b^y = x \).

This equation shows how logs and exponents are inverse operations. By using this conversion, you can transform a logarithmic equation into an exponential equation, which often presents a straightforward arithmetic solution.

For example, given the logarithmic equation \( \log_2\left( \frac{(x+1)(x+5)}{2x+5} \right) = 2 \), converting to exponential form yields \( \frac{(x+1)(x+5)}{2x+5} = 2^2 = 4 \). This step simplifies the problem by eliminating the logarithm and directly solving for the unknown variable.

Quadratic Equations

After transforming the logarithmic problem into an exponential equation and simplifying it, you might encounter a quadratic equation. Recognizing and solving quadratic equations is an essential algebra skill.

A quadratic equation is typically of the form:

• \( ax^2 + bx + c = 0 \)

These equations can be solved by factoring, completing the square, or using the quadratic formula.

In our example, after clearing the fractions and simplifying, we ended up with a quadratic equation: \( x^2 - 2x - 15 = 0 \). This equation can be factored into \( (x - 5)(x + 3) = 0 \). Factoring reveals the potential solutions \( x = 5 \) and \( x = -3 \).

However, in the context of the original problem, not all solutions may be valid due to the nature of logarithms (e.g., negative arguments are not allowed). Thus, it's crucial to validate each solution within the context of the original problem. In this case, \( x = 5 \) is ultimately the only valid solution.