Problem 50 - Algebra and Trigonometry

Question

Solve the logarithmic equation for \(x.\) \(\log _{3}(x+15)-\log _{3}(x-1)=2\)

Step-by-Step Solution

Answer

The solution is \( x = 3 \).

Step 1: Use the Quotient Rule

The equation is given as \( \log_{3}(x+15) - \log_{3}(x-1) = 2 \). According to the properties of logarithms, \( \log_{b}(A) - \log_{b}(B) = \log_{b}\left(\frac{A}{B}\right) \). Apply this rule to combine the logarithms: \[ \log_{3}\left(\frac{x+15}{x-1}\right) = 2. \]

Step 2: Convert Logarithmic to Exponential Form

To remove the logarithm, convert the equation to its exponential form. The equation \( \log_{3}\left(\frac{x+15}{x-1}\right) = 2 \) can be rewritten as \( \frac{x+15}{x-1} = 3^2 \), which simplifies to \( \frac{x+15}{x-1} = 9 \).

Step 3: Clear the Fraction

Multiply both sides of the equation by \( x - 1 \) to eliminate the fraction: \[ x + 15 = 9(x - 1). \]

Step 4: Simplify and Solve for \( x \)

Expand and simplify the equation from the previous step: \[ x + 15 = 9x - 9. \] Rearrange terms to get all \( x \) terms on one side: \[ 15 + 9 = 9x - x. \] This simplifies to \[ 24 = 8x \], so divide by 8: \[ x = 3. \]

Step 5: Verify the Solution

Check the solution by substituting \( x = 3 \) back into the original equation: \( \log_{3}(x+15) - \log_{3}(x-1) = \log_{3}(18) - \log_{3}(2) = \log_{3}\left(\frac{18}{2}\right) = \log_{3}(9) = 2 \). Thus, the solution \( x = 3 \) satisfies the equation.

Key Concepts

Understanding Properties of Logarithms

The properties of logarithms are essential tools for simplifying and solving logarithmic equations. In this exercise, we primarily use the Quotient Rule of logarithms. The Quotient Rule states that for any positive numbers A and B, and a base b, \( \log_{b}(A) - \log_{b}(B) = \log_{b}\left(\frac{A}{B}\right) \).
To apply this rule, we need to identify the logarithms involved and their bases. Here, both terms share the same base of 3. We reformulate the equation by placing the variables in a fraction within a single logarithmic term: \[ \log_{3}\left(\frac{x+15}{x-1}\right) = 2. \]
By understanding and applying the properties of logarithms, complex expressions can often be reduced to simpler forms, making them easier to solve.

Converting to Exponential Form

After applying the properties of logarithms, the next step is to convert the logarithmic equation into an exponential form. This is a crucial step when you need to remove the logarithm from an equation to solve for a variable.
The equation \( \log_{3}\left(\frac{x+15}{x-1}\right) = 2 \) can be rewritten using the definition of a logarithm. The expression maps to an exponential form: \( 3^2 = \frac{x+15}{x-1} \).
Remember, if \( \log_{b}(C) = y \), then in exponential form, it is written as \( b^y = C \). This conversion opens up the path to easily compute values by clearing fractions or isolating the variable involved.

Solving Equations Involving Variables

Once the equation is in exponential form, solving it becomes a straightforward algebraic exercise. In our problem, after converting to an exponential form, it becomes: \( \frac{x+15}{x-1} = 9 \).
To eliminate the fraction, multiply both sides by \( x - 1 \), which yields \( x + 15 = 9(x - 1) \). Solving this expression involves typical algebraic steps. Expand and simplify: \( x + 15 = 9x - 9 \).
Bring all \( x \)-terms to one side: \( 9x - x = 15 + 9 \), leading to \( 8x = 24 \). Dividing both sides by 8 gives \( x = 3 \).
Finally, always substitute your solution back into the original equation for verification. A correct solution will satisfy the initial conditions, confirming its validity.