Problem 47 - Algebra and Trigonometry

Question

Solve the logarithmic equation for \(x.\) \(\log x+\log (x-1)=\log (4 x)\)

Step-by-Step Solution

Answer

The solution is \( x = 5 \).

Step 1: Use Logarithm Property

Combine the logarithms on the left side using the property of logarithms: \( \log a + \log b = \log (a \times b) \). This gives: \( \log(x(x-1)) = \log(4x) \).

Step 2: Apply Logarithm Equality

Since the logarithms are equal, their arguments must be equal. Set the expressions inside the logarithms as an equation: \( x(x-1) = 4x \).

Step 3: Simplify the Equation

Expand the left side of the equation and simplify: \( x^2 - x = 4x \), which simplifies to \( x^2 - x - 4x = 0 \) or \( x^2 - 5x = 0 \).

Step 4: Factor the Equation

Factor the quadratic equation: \( x(x - 5) = 0 \).

Step 5: Solve the Factored Equation

Set each factor to zero and solve for \(x\): \( x = 0 \) or \( x - 5 = 0 \). Therefore, \( x = 0 \) or \( x = 5 \).

Step 6: Verify Solutions

Check the solutions against the original equation. For \( x = 0 \), the logarithm of zero is undefined, so reject this solution. For \( x = 5 \): \( \log 5 + \log 4 = \log 20 = \log(4 \times 5) \) which holds true. Therefore, \( x = 5 \) is a valid solution.

Key Concepts

Properties of Logarithms

Logarithms are powerful tools used in solving equations involving exponential terms. One of the key properties of logarithms is the product property, which states that the logarithm of a product is the sum of the logarithms of the factors. This can be written mathematically as:

• \( \log a + \log b = \log (a \times b) \)

This property is useful when you need to simplify expressions and solve equations like the one in the given exercise. In this problem, we used the product property to combine the logs on the left side:

• \( \log x + \log (x - 1) = \log (x(x - 1)) \)

Remember that these properties are only valid if all arguments of the logarithms are positive. This highlights the importance of checking the domain of the original equation to make sure the solution is valid, especially when variables are involved.
Understanding these properties helps simplify many logarithmic equations efficiently.

Solving Quadratic Equations

After applying logarithmic properties, we transformed the given problem into a quadratic equation. Quadratic equations follow the general form:

• \( ax^2 + bx + c = 0 \)

In our exercise, after simplifying, the quadratic equation becomes:

• \( x^2 - 5x = 0 \)

To solve quadratic equations like this, factoring is one of the primary methods used. Factoring involves expressing the equation as a product of linear terms. Here, we have:

• \( x(x - 5) = 0 \)

To find the roots of the equation, set each factor equal to zero. This gives solutions:

• \( x = 0 \)
• \( x = 5 \)

Always verify the solutions in the context of the original equation to ensure they are valid, as some might not fit due to possible restrictions from earlier steps, like with logarithms.

Logarithm Equality

Logarithms have unique traits when it comes to equality. If you have two expressions where the logarithms of each side are equal, it implies that the arguments themselves must be equal. This concept can be expressed as:

• If \( \log a = \log b \), then \( a = b \)

This property stems from the fact that logarithmic functions are one-to-one, meaning each input corresponds to exactly one output. In our exercise, once the logs are combined using properties, the equation became:

• \( \log(x(x-1)) = \log(4x) \)

This equality allowed us to equate the arguments directly:

• \( x(x-1) = 4x \)

This step is critical in solving logarithmic equations. As it turned the logarithmic equation into a solvable algebraic one. Remember to always check that any solutions derived from this process do not fall outside of the original logarithmic constraints, such as non-negative arguments.