Problem 55 - Elementary and Intermediate Algebra

Question

Solve each equation. See Example \(8 .\) $$ \log _{5}(4 x-1)+\log _{5} x=1 $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{5}{4} \).

Step 1: Simplify Logarithmic Addition

Use the property of logarithms that states \( \log_b{m} + \log_b{n} = \log_b{(mn)} \). Apply this to combine the logarithms: \[ \log_{5}((4x-1)x) = 1 \] This simplifies the equation to a single logarithm.

Step 2: Exponential Form Transformation

Convert the logarithmic equation to its exponential form. Since \( \log_{5}(A) = B \) can be rewritten as \( 5^B = A \), our equation becomes: \[ 5^1 = (4x-1)x \] This simplifies to: \[ 5 = 4x^2 - x \]

Step 3: Formulate a Quadratic Equation

Rearrange the quadratic equation by moving all terms to one side: \[ 4x^2 - x - 5 = 0 \] This forms a standard quadratic equation \( ax^2 + bx + c = 0 \).

Step 4: Solve the Quadratic Equation

Apply the quadratic formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \) with \( a = 4 \), \( b = -1 \), \( c = -5 \): \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-5)}}{2(4)} \]Simplify under the square root and solve: \[ x = \frac{1 \pm \sqrt{1 + 80}}{8} \] \[ x = \frac{1 \pm 9}{8} \]Thus, \( x = \frac{10}{8} = \frac{5}{4} \) or \( x = \frac{-8}{8} = -1 \).

Step 5: Determine Valid Solutions

Evaluate these solutions in the context of the original logarithmic equation. 1. \( x = \frac{5}{4} \) is valid as \( 4x-1 = 4(\frac{5}{4})-1 = 4 \), giving valid logs. 2. \( x = -1 \) is invalid as logarithms of negative numbers are undefined.Thus, the only valid solution is \( x = \frac{5}{4} \).

Key Concepts

Properties of Logarithms

Logarithms may initially seem complex, but they become much simpler once you understand their properties. One of the most important properties is the product rule, which states that the sum of two logarithms with the same base is equivalent to the logarithm of the product of their arguments. Mathematically, this can be expressed as:

• \( \log_b{m} + \log_b{n} = \log_b{(mn)} \)

In the exercise, we applied this property to combine \( \log_{5}(4x-1) \) and \( \log_{5}x \) into a single logarithm:

• \( \log_{5}((4x-1)x) = 1 \)

This property helps simplify the equation, making it easier to work with in further steps. Remember, such rules can only be applied when the logarithms share the same base.
If the property sounds confusing, think of it as a tool to simplify complex equations. By harnessing this property, we reduce the complexity and reveal a path to solutions.

Quadratic Equations

Quadratic equations often arise when simplifying logarithmic equations through exponential transformations. A quadratic equation follows the standard form:

• \( ax^2 + bx + c = 0 \)

In our problem, when we transformed the logarithmic equation to an exponential form, we ended up with the quadratic equation:

• \( 4x^2 - x - 5 = 0 \)

To solve this equation, we used the quadratic formula:

• \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \)

Substituting \( a = 4 \), \( b = -1 \), and \( c = -5 \), we found two potential solutions for \( x \):
\[ x = \frac{5}{4} \quad \text{and} \quad x = -1 \]
Quadratic equations can seem daunting, but by following step-by-step procedures like this, you can systematically work through the problem and find all potential solutions.

Exponential Form Transformation

Transforming a logarithmic equation into its exponential form is a key step in solving many equations involving logs. The essence of this transformation is to switch from an equation based on logarithms to one based on exponents, which can often simplify the problem.

• The basic concept is expressed as: if \( \log_b(A) = C \), then \( b^C = A \).

In our problem, we had the equation:

• \( \log_{5}((4x-1)x) = 1 \)

Using the property mentioned, we transformed this into:

• \( 5^1 = (4x-1)x \)

This transformation allowed us to step away from the logarithmic form and directly address the algebraic expression, further simplifying the path to solving for \( x \). Remember, converting to exponential form can drastically open up new avenues for solving an equation and is an invaluable tool in algebra.