Problem 47 - Precalculus

Question

Solve the logarithmic equations exactly. $$\log _{5}(x-4)+\log _{5} x=1$$

Step-by-Step Solution

Answer

The solution to the equation is \( x = 5 \).

Step 1: Use the Product Property of Logarithms

Recall that the product property of logarithms states that \( \log_b (M) + \log_b (N) = \log_b (M \times N) \). Thus, we can combine the terms: \( \log_5((x-4) \times x) = 1 \).

Step 2: Convert the Equation to Exponential Form

Use the definition of a logarithm to convert the equation to its exponential form. Since \( \log_b(A) = C \) is equivalent to \( b^C = A \), we have \( 5^1 = (x-4) \times x \).

Step 3: Simplify and Set Up a Quadratic Equation

The exponential form gives us the equation \( 5 = x^2 - 4x \). Rearrange this to form a standard quadratic equation: \( x^2 - 4x - 5 = 0 \).

Step 4: Factor the Quadratic Equation

To solve the quadratic equation \( x^2 - 4x - 5 = 0 \), we can factor it. Find two numbers that multiply to \(-5\) and add to \(-4\). These numbers are \(-5\) and \(1\). Therefore, \( (x-5)(x+1) = 0 \).

Step 5: Solve for x

Set each factor equal to zero and solve for \(x\): 1. \( x - 5 = 0 \Rightarrow x = 5 \) 2. \( x + 1 = 0 \Rightarrow x = -1 \).

Step 6: Check for Logarithm Domain Restrictions

Check the solutions against the domain restrictions of the logarithms. For both logarithms to be defined, \( x-4 > 0 \) and \( x > 0 \), implying \( x > 4 \). Thus, only \( x = 5 \) is valid since \( x = -1 \) does not satisfy these conditions.

Key Concepts

Product Property of Logarithms

The Product Property of Logarithms is a useful tool when dealing with logarithmic expressions. This property allows us to combine two logarithms with the same base into a single logarithm. Specifically, it states that:

• If you have two logarithms,
  \( \log_b(M) + \log_b(N) \),
  you can combine them as \( \log_b(M \times N) \).

In the given problem, this property is applied to \( \log_5(x - 4) + \log_5(x) \) to form \( \log_5((x - 4) \times x) \). This simplification is crucial because it reduces the complexity of the equation, making it easier to solve. Next, we'll see how to convert this equation using exponential form.

Exponential Form

The exponential form of a logarithmic equation is derived from the definition of a logarithm. If you have a logarithmic expression like \( \log_b(A) = C \),
this is equivalent to writing the exponential form as \( b^C = A \).In the exercise, we transform \( \log_5((x - 4) \times x) = 1 \) into its exponential form:
\( 5^1 = (x - 4) \times x \).

• This step makes it easier to find the numerical solutions.
  It enables moving from a logarithmic framework to a more straightforward polynomial one.

The solution becomes more tractable as the exponential form opens the path to further simplifications, leading to a quadratic equation.

Quadratic Equations

A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where a, b, and c are constants. These equations are called quadratic because their highest exponent for the variable x is 2.Upon converting to exponential form in our problem, we arrive at the equation:
\( 5 = x^2 - 4x \).
After rearranging terms, it becomes \( x^2 - 4x - 5 = 0 \).To solve this, we typically factor the quadratic expression, if possible. Here, we find two numbers that multiply to -5 and add up to -4, which are \(-5\) and \(1\). Consequently, the equation factors to:

• \((x - 5)(x + 1) = 0\).

By setting each factor equal to zero, we find the potential solutions to the quadratic equation.

Domain Restrictions

Domain restrictions in logarithmic functions are crucial because logarithms are not defined for zero or negative numbers. Thus, when solving logarithmic equations, it is important to check whether the solutions fall within the permissible domain.For the exercise, the domain restrictions are

• \(x - 4 > 0\) and
• \(x > 0\), which imply \(x > 4\).

Evaluating the solutions \(x = 5\) and \(x = -1\) against these restrictions, we find that only \(x = 5\) satisfies the conditions.
Thus, \(x = 5\) is the valid solution because \(x = -1\) leads to negative or undefined values in the original logarithmic expressions.
Ensuring solutions respect domain restrictions guarantees that they are mathematically and practically valid.