Problem 55 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The solution is \( x = \frac{5}{4} \). \( x = -1 \) is not valid.

Step 1: Apply Logarithmic Product Rule

The first step is to use the logarithmic property that states \( \log_a b + \log_a c = \log_a (bc) \). Here, it will combine \( \log_{5}(4x - 1) \) and \( \log_{5}x \) into a single logarithm: \[ \log_{5}((4x - 1) x) = 1 \] This simplifies to:\[ \log_{5}(4x^2 - x) = 1 \]

Step 2: Convert to Exponential Form

Now, we need to convert the logarithmic equation to exponential form. The equation \( \log_{5}(4x^2 - x) = 1 \) implies:\[ 5^1 = 4x^2 - x \] Thus, simplifying gives:\[ 4x^2 - x = 5 \]

Step 3: Rearrange into Quadratic Form

Reorganize the equation to present it as a standard quadratic equation:\[ 4x^2 - x - 5 = 0 \]

Step 4: Solve the Quadratic Equation

The quadratic equation \( 4x^2 - x - 5 = 0 \) can be solved using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = -1 \), \( c = -5 \). Calculate the discriminant first:\[ b^2 - 4ac = (-1)^2 - 4(4)(-5) = 1 + 80 = 81 \]Apply the quadratic formula:\[ x = \frac{-(-1) \pm \sqrt{81}}{2(4)} = \frac{1 \pm 9}{8} \]Solving gives two potential solutions:\( x = \frac{1 + 9}{8} = \frac{10}{8} = \frac{5}{4} \)\( x = \frac{1 - 9}{8} = \frac{-8}{8} = -1 \)

Step 5: Validate Solutions

We need to check whether the solutions are valid for the original logarithmic equation. Recall that the argument in a logarithm must be positive.For \( x = \frac{5}{4} \):- Check \( \log_{5}(4x - 1) \) : \( 4(\frac{5}{4}) - 1 = 5 - 1 = 4 > 0 \)- Check \( \log_{5}(x) \) : \( \log_{5}(\frac{5}{4}) \) where \( \frac{5}{4} > 0 \)Both are positive, so \( x = \frac{5}{4} \) is valid.For \( x = -1 \):- \( \log_{5}(-1) \) is undefined as the logarithm must have a positive argument.Thus, \( x = -1 \) is not valid.

Key Concepts

Logarithmic Properties

Logarithmic properties are essential tools for solving complex equations involving logarithms. A critical rule here is the **Product Rule for Logarithms** which states that the sum of two logarithms with the same base can be expressed as a single logarithm of the product of their arguments:

• Formula: \(\log_a b + \log_a c = \log_a (bc)\)
• Example: \(\log_5(4x - 1) + \log_5 x = \log_5((4x - 1)x)\)

This property simplifies several parts of equations, making them easier to tackle through further algebraic steps.
The base of the logarithm remains unchanged while combining the arguments into a single logarithmic term. This simplification often converts a complex problem into a more manageable form.

• Remember, logarithms require positive arguments, meaning each part inside a log must be greater than zero.
• This ensures that the operations defined are mathematically valid.

Using these properties effectively can unravel complicated mathematical problems, bringing clarity to confusing logarithmic expressions.

Quadratic Formulas

Once a logarithmic equation is simplified and converted using properties, it often takes either a linear or quadratic form. For quadratic forms, solving the equation involves identifying the coefficients and applying the **Quadratic Formula**.

• Formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
• Where \(a\), \(b\), and \(c\) are constants in the quadratic equation \(ax^2 + bx + c = 0\).

In our problem, the quadratic equation \(4x^2 - x - 5 = 0\) is solved this way:

• Calculate the discriminant: \(b^2 - 4ac\).
• Check if it's positive, zero, or negative to identify the nature of roots.
• Substitute values into the quadratic formula to find \(x\).

Understanding the quadratic formula is crucial because it arguably becomes an inflexible tool for solving higher-degree polynomial equations when other simpler methods don't apply. It also offers a method to determine the limitations and nature of solutions, such as real or imaginary, which are key in fully grasping the solutions of quadratic-like structures in equations.

Exponential Equations

When working with logarithms, converting to an exponential form is often an effective strategy to isolate and solve for unknowns. An **Exponential Equation** is one where terms involving the variable are in the form of a base raised to a power involving the variable.

• Conversion: If \(\log_a (x) = b\), then \(x = a^b\).
• Example: \(\log_5 (4x^2 - x) = 1\) becomes \(4x^2 - x = 5^1\).

This step is pivotal in translating logarithmic equations into a form that can be further manipulated through standard algebraic techniques.
The transition from logarithmic to exponential format simplifies the structure of equations. It allows the elimination of logarithmic terms, displaying the core polynomial or equation structure immediately accessible to other solving skills such as factoring or using the quadratic formula.
Mastering the art of switching between logarithmic and exponential representations is a valuable skill, aiding in understanding and solving more complex mathematical problems efficiently.