Problem 72 - 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. $$ \log _{4}(x+2)-\log _{4}(x-1)=1 $$

Step-by-Step Solution

Answer

The solution to the logarithmic equation is \(x = 2\).

Step 1: Apply logarithm laws

Firstly, apply the properties of logarithms to combine the two logarithmic expressions on the left side of the equation. According to the laws of logarithms, the difference of two logarithms is equivalent to the logarithm of the quotient. This gives us: \[ \log _{4} \left(\frac{x+2}{x-1}\right) = 1 \]

Step 2: Convert logarithm to exponential form

The equation is now in the form \(\log_{b} a = c\), which can be converted into its equivalent exponential form \(b^c = a\). Therefore, the equation becomes: \[ 4^1 = \frac{x+2}{x-1} \] Simplifying this equation gives us: \[ 4 = \frac{x+2}{x-1} \]

Step 3: Solve for x

Next, solve for \(x\) by multiplying both sides of the equation by \(x - 1\) and then simplifying. We get: \(4(x-1) = x+2 \rightarrow 4x - 4 = x + 2 \rightarrow 3x = 6 \) This equation simplifies to \(x = 2\).

Step 4: Check the solution

The final step is to check the solution against the original logarithmic expressions to ensure the solution falls into the allowable domain. Substitute \(x = 2\) into the original logarithmic expressions and check if the expressions are defined. We get \(2 + 2 = 4 > 0\) and \(2 - 1 = 1 > 0\). Hence, \(x = 2\) is a valid solution.

Key Concepts

Properties of Logarithms

Understanding the properties of logarithms is essential when working with logarithmic equations. These properties help us to manipulate and solve expressions involving logarithms. One of the most useful properties is the quotient rule:

• Quotient Rule: The logarithm of a quotient is equal to the difference of the logarithms: \( \log_b \left( \frac{A}{B} \right) = \log_b A - \log_b B \).

This rule allows us to combine two logarithms into one, simplifying the solving process. For example, the original problem \( \log _{4}(x+2)-\log _{4}(x-1) = 1 \) was transformed using this property into \( \log _{4} \left(\frac{x+2}{x-1}\right) = 1 \).
By rewriting the equation, we make it easier to solve by converting it into exponential form. Always remember these properties as they're key tools for handling logarithmic equations.

Exponential Form Conversion

Converting logarithmic equations to their equivalent exponential forms is a crucial step in solving them. The general formula to remember is:

• If \( \log_{b} a = c \), then \( b^c = a \).

This formula translates a logarithmic problem into an exponential one, which is often easier to solve. In our initial equation, after applying the properties of logarithms, we arrived at \( \log _{4} \left(\frac{x+2}{x-1}\right) = 1 \).
To convert this, we used the formula to get the equivalent exponential equation: \( 4^1 = \frac{x+2}{x-1} \). This step is integral as it paves the way for straightforward algebraic manipulation, allowing us to isolate and solve for \(x\).
Exponential conversion is like a bridge that transforms complex logarithmic problems into more familiar algebraic equations.

Solving Equations

Solving equations involves using algebraic techniques to isolate the variable of interest. Once the logarithmic equation \( 4 = \frac{x+2}{x-1} \) is obtained from exponential form conversion, our goal is to solve for \(x\).

Here's a step-by-step approach:

• Multiply both sides by the denominator \((x-1)\) to eliminate the fraction: \(4(x-1) = x+2\).
• Distribute and simplify: \(4x - 4 = x + 2\).
• Isolate \(x\) by moving terms: \(4x - x = 2 + 4\) resulting in \(3x = 6\).
• Finally, divide by 3 to find \(x = 2\).

By following these steps, you can efficiently find the value of \(x\). It’s important to work carefully through each step, ensuring that all manipulations are valid and correct.

Domain Restrictions

When solving logarithmic equations, it's critical to consider the domain of the logarithmic expressions. Logarithms are defined only for positive real numbers, so it's paramount to ensure that any value of \(x\) results in positive arguments within the logarithms.

For the equation \( \log _{4}(x+2)-\log _{4}(x-1) = 1 \), we need to check:

• \(x+2 > 0\), which simplifies to \(x > -2\).
• \(x-1 > 0\), which simplifies to \(x > 1\).

The tighter restriction here is \(x > 1\), which must be satisfied for any solution to be valid. When you solve the equation and find \(x = 2\), you need to verify that it respects this condition. Substituting back confirms both \(x+2\) and \(x-1\) are positive, ensuring our solution is within the valid domain.
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