Problem 79 - Algebra and Trigonometry

Question

Solve each logarithmic equation. Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ \log (3 x-3)=\log (x+1)+\log 4 $$

Step-by-Step Solution

Answer

The equation has no solution.

Step 1: Combine the Logarithms

Use the property \(\log a + \log b = \log (ab)\) to rewrite the right-hand side of the equation: \(\log (3x - 3) = \log ((x + 1) * 4)\). Now we have \(\log (3x - 3) = \log (4x + 4)\). According to the one-to-one property of logarithms, if \(\log a = \log b\), then \(a = b\). Therefore, we can form a new equation: \(3x - 3 = 4x + 4\).

Step 2: Solve for x

Isolate x by subtracting \(4x\) from both sides to get \(-x - 3 = 4\), and then add 3 to both sides to get \(-x = 7\). Multiply both sides by -1 to solve for \(x\): \(x = -7\).

Step 3: Verify the Solution

We have to make sure that this solution is in the domain of the original logarithmic expressions. This would mean that \(3x - 3 > 0\) and \(x + 1 > 0\). For \(x = -7\), both inequalities are not true. Thus, the solution \(x = -7\) is not in the domain of the original logarithmic expressions and should be rejected.

Step 4: No Solution

The proposed solution is not suitable, hence this equation has no solution.

Key Concepts

Logarithm Properties

Understanding logarithm properties is essential when solving logarithmic equations. A logarithm, denoted as \( \log_b a \), is the power to which the base \( b \) must be raised to obtain the number \( a \). Here are some key properties:

• The Product Rule: \( \log_b (MN) = \log_b M + \log_b N \), which allows us to add two logarithms with the same base.
• The Quotient Rule: \( \log_b \left(\frac{M}{N}\right) = \log_b M - \log_b N \), which lets us subtract two logarithms with the same base.
• The Power Rule: \( \log_b (M^k) = k \cdot \log_b M \), expressing that a logarithm of a power allows the exponent to be brought to the front.

These rules simplify logarithmic expressions and solving equations. For instance, during the step by step solution of the exercise, we combined the terms on the right using the Product Rule: \( \log (x+1) + \log 4 = \log ((x + 1) * 4) \). It allowed us to set up the equation for a simpler comparison, leading us closer to finding the solution.

Domain of Logarithmic Functions

The domain of logarithmic functions is another crucial concept when solving logarithmic equations. The domain includes all the possible values of \( x \) that make the function real and defined. For any logarithmic function \( \log_b (x) \), the argument \( x \) must be greater than zero because you cannot take the logarithm of a negative number or zero in the real number system.

• The requirement is \( x > 0 \), ensuring the expression inside the logarithm is positive.

When analyzing our exercise solution, we have to check whether the proposed values for \( x \) make the logarithmic expressions real and defined. After we have the potential solution of \( x = -7 \), we must test if this value satisfies \( 3x - 3 > 0 \) and \( x + 1 > 0 \). Since both conditions fail for \( x = -7 \), the value is not in the domain of the original logarithmic functions, hence rejected.

One-to-One Property of Logarithms

The one-to-one property of logarithms is pivotal for solving equations that contain logarithmic terms. This property states that if \( \log_b (M) = \log_b (N) \), then \( M = N \). This property is derived from the fact that a logarithmic function is one-to-one—or in other words, it does not repeat values for different inputs.

Practical Use in Equations

When faced with logarithmic equations, like in our exercise, we use this property to set the terms inside the logarithms equal to each other once their bases are the same and their logarithms are equal. In the step by step solution provided, after combining the logs, we ended up with \( \log (3x - 3) = \log (4x + 4) \). Invoking this one-to-one property allowed us to deduce that \( 3x - 3 = 4x + 4 \), greatly simplifying our problem. However, the resulting value must still fall within the domain of the original logarithmic functions to be considered a valid solution.