Problem 48 - College Algebra

Question

For the following exercises, solve each equation for \(x\). $$ \log _{8}(x+6)-\log _{8}(x)=\log _{8}(58) $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{2}{19} \).

Step 1: Apply the Logarithm Quotient Rule

The equation given is \( \log_{8}(x+6) - \log_{8}(x) = \log_{8}(58) \). Use the logarithm quotient rule, which states that \( \log_{b}(A) - \log_{b}(B) = \log_{b}\left(\frac{A}{B}\right) \), to combine the logs on the left side:\[ \log_{8}\left(\frac{x+6}{x}\right) = \log_{8}(58) \]

Step 2: Set the Arguments Equal

Since the logarithms are equal and have the same base, their arguments must also be equal. Set the arguments equal to each other:\[ \frac{x+6}{x} = 58 \]

Step 3: Clear the Fraction

To remove the fraction, multiply both sides of the equation by \(x\):\[ x+6 = 58x \]

Step 4: Rearrange the Equation

Rearrange the equation to get terms involving \(x\) on one side:\[ x - 58x = -6 \]Combine like terms:\[ -57x = -6 \]

Step 5: Solve for x

Solve the equation by dividing both sides by \(-57\):\[ x = \frac{-6}{-57} = \frac{6}{57} \]Simplify the fraction:\[ x = \frac{2}{19} \]

Step 6: Verify the Solution

Plug \( x = \frac{2}{19} \) back into the original equation to verify it satisfies the equation:- Calculate \( \log_{8}\left(\frac{2}{19} + 6\right) \) and \( \log_{8}\left(\frac{2}{19}\right) \).- Check that their difference equals \( \log_{8}(58) \).After verification, \( x = \frac{2}{19} \) is indeed a valid solution.

Key Concepts

logarithm quotient rule

The logarithm quotient rule is a useful property in simplifying expressions involving logs. This rule allows you to combine or simplify the subtraction of two logarithms with the same base.For instance, if you have an expression like \( \log_b(A) - \log_b(B) \), the rule states that it can be rewritten as \( \log_b\left( \frac{A}{B} \right) \).This is applicable only when the base \(b\) of the logs and the expression inside the logs \((A)\) are the same. In our given problem:

• We applied this to the left side: \( \log_{8}(x+6) - \log_{8}(x) = \log_{8}\left( \frac{x+6}{x} \right) \).
• This step was important to combine the logs, simplifying the equation and making it easier to solve.

Understanding the quotient rule helps you clear complexities when dealing with logarithms, making your calculations faster and minimizing errors.

solving equations

In equations involving logarithms, like the one in our example, solving for \( x \) often requires algebraic manipulation after using logarithmic properties. Once the logarithms have been combined using the quotient rule:

• We then set the arguments equal: \( \frac{x+6}{x} = 58 \), as the left side expression equals the right side in the log form.
• To eliminate the fraction, multiply both sides by \(x\). This yields \( x + 6 = 58x \).
• Rearrange this to get all terms involving \(x\) on one side, making it ready for a straightforward solution: \( x - 58x = -6 \). Combining terms, you find \( -57x = -6 \).
• Finally, divide both sides by \(-57\) to isolate \(x\), getting \( x = \frac{6}{57} \), which simplifies to \( x = \frac{2}{19} \).

Each step moves you closer to finding the variable by simplifying the equation through multiplication, division, or subtraction. These operations are essential in solving similar algebraic equations.

verification of solutions

Verification is an indispensable part of solving equations, especially those involving logarithms. It ensures that the solution is correct and satisfies the original equation.Here's how verification is done for our problem:

• Once we found \( x = \frac{2}{19} \), substitute \( x \) back into the original equation \( \log_{8}(x+6) - \log_{8}(x) = \log_{8}(58) \).
• Calculate the expression \( \log_{8}\left( \frac{2}{19} + 6 \right) \) and \( \log_{8}\left( \frac{2}{19} \right) \).
• Their difference should equal \( \log_{8}(58) \)

This step checks if the derived \(x\) value, when used in the original setup, holds true and does not lead to contradictions or undefined expressions. Verification confirms the correctness and completeness of the solution obtained.