Problem 39 - College Algebra

Question

Use the quotient rule for logarithms to find all \(\mathrm{x}\) values such that \(\log _{6}(x+2)-\log _{6}(x-3)=1 .\) Show the steps for solving.

Step-by-Step Solution

Answer

The solution is \( x = 4 \).

Step 1: Apply the Quotient Rule for Logarithms

The quotient rule of logarithms states that \( \log_b(\frac{M}{N}) = \log_b(M) - \log_b(N) \). We can use this rule to combine the logarithms on the left-hand side: \( \log_6(x+2) - \log_6(x-3) = \log_6(\frac{x+2}{x-3}) \). The equation now becomes: \[ \log_6\left(\frac{x+2}{x-3}\right) = 1. \]

Step 2: Exponentiate Both Sides

To eliminate the logarithm, convert the logarithmic equation to its exponential form. Since \( \log_6\left(\frac{x+2}{x-3}\right) = 1 \), rewrite this as \( \frac{x+2}{x-3} = 6^1 \). Thus, \( \frac{x+2}{x-3} = 6 \).

Step 3: Solve the Rational Equation

To solve \( \frac{x+2}{x-3} = 6 \), multiply both sides by \( x-3 \) to clear the fraction: \( x+2 = 6(x-3) \). Distribute the 6 on the right-hand side: \( x+2 = 6x - 18 \).

Step 4: Isolate x

Subtract \( x \) from both sides to get \( 2 = 5x - 18 \). Add 18 to both sides: \( 2 + 18 = 5x \). That simplifies to \( 20 = 5x \). Divide both sides by 5 to solve for \( x \): \( x = 4 \).

Step 5: Verify the Solution

Substitute \( x = 4 \) back into the original argument of each logarithm to ensure they are within their domains: \( \log_6(4+2) - \log_6(4-3) = \log_6(6) - \log_6(1) \). Since \( \log_6(1) = 0 \), the equation becomes \( \log_6(6) = 1 \), confirming that \( x = 4 \) is a valid solution.

Key Concepts

Quotient Rule

The quotient rule in logarithms is an essential tool for simplifying expressions involving logs. It allows us to combine two logs with the same base that are being subtracted. The rule states:

• \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \)

This means if you have a difference of two logs, you can rewrite them as a single log of the division of their arguments. In our exercise, we see:

• \( \log_6(x+2) - \log_6(x-3) \)

Applying the quotient rule, this becomes:

• \( \log_6\left(\frac{x+2}{x-3}\right) \)

This simplification is crucial because it allows us to set up the equation for solving by reducing complexity.

Exponential Form

Converting a logarithmic equation to exponential form is a common technique when solving for variables. The core idea is that every logarithmic expression can be switched to an exponential expression. Given the equation:

• \( \log_b(A) = C \)

It translates to:

• \( A = b^C \)

In our problem, after using the quotient rule, we have:

• \( \log_6\left(\frac{x+2}{x-3}\right) = 1 \)

Switching it to exponential form gives:

• \( \frac{x+2}{x-3} = 6^1 \)

This step effectively eliminates the logarithm, bringing the problem into a simple equation we can solve step-by-step.

Solving Rational Equations

Rational equations involve fractions where the numerator and/or the denominator contains a variable. Solving them often involves clearing the fractions by multiplying through by an appropriate value. In our exercise, after converting to exponential form, we have the equation:

• \( \frac{x+2}{x-3} = 6 \)

To clear the fraction, multiply both sides by \(x-3\):

• \( x+2 = 6(x-3) \)

Expanding the right side yields:

• \( x+2 = 6x - 18 \)

By rearranging the terms and solving, we isolate \(x\) to find the value satisfying the equation.
Always remember to check your solution in the original rational expressions to ensure no division by zero or incorrect domains.

Domain of Logarithms

When working with logarithmic functions, understanding their domain is vital. The domain specifies all the possible input values for the variable that make the logarithmic expression valid. For any log function \( \log_b(M) \), the argument \(M\) must satisfy:

• \( M > 0 \)

This requirement stems from the fact that you can't take the logarithm of zero or a negative number.
In the context of our exercise, the original logarithms were \( \log_6(x+2) \) and \( \log_6(x-3) \). This means the conditions for the domain should be:

• \( x+2 > 0 \Rightarrow x > -2 \)
• \( x-3 > 0 \Rightarrow x > 3 \)

Thus, the intersection of these conditions \( x > 3 \) ensures that the argument of every log is positive. It's a crucial step to confirm that our solution lies within this domain to be valid.