Problem 84 - College Algebra and Calculus: An Applied Approach

Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\log _{3}(x+8)=\log _{3}(3 x+2)\)

Step-by-Step Solution

Answer

The solution of the logarithmic equation is \(x = 3\).

Step 1: Set the arguments of the logarithms equal to each other

Since the logs on both sides of the equation have the same base, we can equate their arguments. This gives us the equation \(x+8 = 3x + 2\).

Step 2: Solve the algebraic equation

Rearrange the equation by subtracting \(x\) from both sides to get \(8 = 2x +2\). Then subtract 2 from both sides to get \(6 = 2x\). Finally, divide both sides by 2 to solve for \(x\), yielding \(x = 3\). Note that when solving logarithmic equations, we must check the solution to ensure that it doesn't result in a log of a negative number, as this isn't defined. In this case, substituting \(x = 3\) back in to our original logarithms does not result in a log of a negative number, so \(x = 3\) is a valid solution.

Step 3: checking the solution

To check the solution, substitute \(x = 3\) back into the original equation. If the left side equals the right side, then the solution is correct. Here the original equation is \(\log _{3}(x+8)=\log _{3}(3 x+2)\). By substituting we get \(\log _{3}(3+8)=\log _{3}(3*3+2)\) which simplifies to \(\log _{3}(11)=\log _{3}(11)\). Since the left side of the equation equals the right side, our solution \(x = 3\) is verified.

Key Concepts

Algebraic Solutions

Algebraic solutions involve solving equations using straightforward algebraic methods to find an unknown value. In our case, we have a logarithmic equation, which initially might seem complicated. However, by applying basic algebraic rules, it transforms into something more manageable.
In the given equation, \( \log_{3}(x+8) = \log_{3}(3x+2) \), we notice that both sides have the same logarithm base. This allows us to use the property that if \( \log_b(a) = \log_b(b) \), then \( a = b \).
Hence, we can set the expressions \( x + 8 \) and \( 3x + 2 \) equal to each other and solve the resulting algebraic equation:

• Rearrange the terms: \( x+8 = 3x+2 \).
• Simplify by isolating \( x \): Subtract \( x \) from both sides, resulting in \( 8 = 2x + 2 \).
• Further simplify by subtracting 2: \( 6 = 2x \).
• Lastly, divide by 2 to find \( x = 3 \).

This step-by-step breakdown highlights the power of algebraic manipulation in solving equations, even when logarithms are involved.

Checking Solutions

Verifying your solution is crucial, especially with logarithmic equations, where values must stay within certain constraints.
Logarithms are only defined for positive numbers, meaning once you solve the equation, it's important to ensure that inserting your solution back into the equation doesn't lead to taking the logarithm of a negative number.
To check our solution \( x = 3 \), we substitute it back into the original logarithmic equation:

• Original equation: \( \log_{3}(x+8) = \log_{3}(3x+2) \).
• Substitute \( x = 3 \): \( \log_{3}(3+8) = \log_{3}(3 \times 3 + 2) \).
• Simplifies to: \( \log_{3}(11) = \log_{3}(11) \).

Since both sides are equal, the solution \( x = 3 \) satisfies the equation, confirming it's correct and valid. Always follow this validating step to ensure your solutions are accurate.

Logarithmic Functions

Understanding logarithmic functions is essential for solving equations involving logarithms. A logarithm answers the question: 'To what power must the base \( b \) be raised to obtain a certain number?'
For instance, \( \log_3(9) = 2 \) because \( 3^2 = 9 \). In a logarithmic function \( \log_b(x) \), \( b \) is the base, and \( x \) is the number we're taking the logarithm of.
Logarithmic functions have unique properties that simplify solving equations:

• \( \log_b(mn) = \log_b(m) + \log_b(n) \), which can simplify multiplications.
• \( \log_b(\frac{m}{n}) = \log_b(m) - \log_b(n) \), useful for divisions.
• \( \log_b(m^n) = n \log_b(m) \), enables handling exponents easily.

These properties are extremely useful in algebraic solutions for converting complex expressions into simpler, solvable forms. Understanding these concepts allows for a clearer strategy when tackling logarithmic equations.