Problem 50 - Precalculus Mathematics for Calculus

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: Apply Logarithm Properties

Use the properties of logarithms to combine the terms on the left side of the equation. The property \(\log_b(m) - \log_b(n) = \log_b\left(\frac{m}{n}\right)\) allows us to write the equation as \(\log_{3}\left(\frac{x+15}{x-1}\right) = 2\).

Step 2: Convert to Exponential Form

To remove the logarithm, convert the logarithmic equation into its exponential form. The equation \(\log_b(y) = n\) can be rewritten as \(b^n = y\). Thus, \(3^2 = \frac{x+15}{x-1}\).

Step 3: Solve the Exponential Equation

Evaluate \(3^2\) to get 9, so the equation becomes \(9 = \frac{x+15}{x-1}\). Cross-multiply to clear the fraction: \(9(x-1) = x+15\).

Step 4: Simplify and Solve for x

Expand and simplify the equation from Step 3: \(9x - 9 = x + 15\). Rearrange terms to get \(8x = 24\), and then divide both sides by 8 to solve for \(x\): \(x = 3\).

Step 5: Verify the Solution

Check the solution by substituting \(x = 3\) back into the original equation to ensure it holds true. Substituting, we have \(\log_{3}(3+15) - \log_{3}(3-1) = 2\), which simplifies to \(\log_{3}(18) - \log_{3}(2) = 2\). This holds since \(\log_{3}(9) = 2\), confirming \(x = 3\) is correct.

Key Concepts

Properties of Logarithms

When dealing with logarithmic equations, understanding the properties of logarithms can be a powerful tool. One such property is the subtraction rule: if you have \( \log_b(m) - \log_b(n) \), you can combine them into a single logarithm: \( \log_b\left(\frac{m}{n}\right) \). This property is derived from the definition of logarithms.
When you see this rule in action, it helps simplify complex equations by reducing the number of terms involved.

• This is especially useful in solving equations, where isolating the variable is key.
• Using this rule can convert a subtraction problem into a division, which is often easier to manage.

So, if you encounter an equation like \( \log_{3}(x+15)-\log_{3}(x-1)=2 \), applying this property allows you to streamline your work and makes further manipulation of the equation simpler.

Exponential Form

Once you have used logarithm properties to simplify an equation, the next natural step is converting the logarithmic equation into exponential form. This involves rearranging the equation into a form without logarithms.
The conversion rule \( \log_b(y) = n \) translates to \( b^n = y \). This is a direct relationship between powers and logs, providing a straightforward path to rid equations of logs.

• Transforming to exponential form can clarify the solution path by simplifying the logical structure of the equation.
• With exponential equations, your variable may be more straightforward to isolate and solve for.

For example, the equation \( \log_{3}\left(\frac{x+15}{x-1}\right) = 2 \) can be converted to \( 3^2 = \frac{x+15}{x-1} \). With the exponential form, your variable expressions are clearer, simplifying the mechanics of solving the equation.

Solving Equations

Once equations are in exponential form, solving them becomes a procedural set of steps. The key is isolating the variable, typically involving simplifying and solving a linear equation. After you have an exponential equation like \( 3^2 = \frac{x+15}{x-1} \), first, you calculate \( 3^2 \), which equals 9.
Then, clear the fraction by cross-multiplying: \( 9(x-1) = x+15 \). Distributing the 9 gives \( 9x - 9 = x + 15 \), which you then rearrange and simplify to solve for \( x \).

• By getting all \( x \) terms on one side, the math simplifies to a linear equation.
• After arranging, solve by isolating \( x \) through addition, subtraction, multiplication, or division.

This process finally leads to \( x = 3 \). Always remember to verify your solution by substituting it back into the original equation, ensuring consistency and correctness.