Problem 36 - Algebra 2

Question

Solve each equation. Check your solutions. \(\log _{2}(y+2)-\log _{2}(y-2)=1\)

Step-by-Step Solution

Answer

The solution is \( y = 6 \).

Step 1: Understand the properties of logarithms

To solve the equation \( \log_{2}(y+2)-\log_{2}(y-2)=1 \), start by recalling the logarithmic property that \( \log_{b}(A) - \log_{b}(B) = \log_{b} \left( \frac{A}{B} \right) \). This allows us to write the equation as \( \log_{2} \left( \frac{y+2}{y-2} \right) = 1 \).

Step 2: Rewrite the equation in exponential form

Because \( \log_{b}(x) = c \) means that \( x = b^c \), the equation \( \log_{2} \left( \frac{y+2}{y-2} \right) = 1 \) can be rewritten as \( \frac{y+2}{y-2} = 2^1 \). This simplifies to \( \frac{y+2}{y-2} = 2 \).

Step 3: Clear the fraction by solving the equation

Multiply both sides of the equation \( \frac{y+2}{y-2} = 2 \) by \( y-2 \) to eliminate the fraction: \( y+2 = 2(y-2) \).

Step 4: Distribute and simplify

Distribute the 2 on the right side: \( y+2 = 2y - 4 \).

Step 5: Isolate the variable y

Move \( y \) from the right side to the left: \( y+2 - 2y = -4 \). Simplify: \( -y + 2 = -4 \).

Step 6: Solve for y

Subtract 2 from both sides: \( -y = -6 \). Multiply both sides by -1: \( y = 6 \).

Step 7: Check the solution in the original equation

Substitute \( y = 6 \) back into the original equation: \( \log_{2}(6+2)-\log_{2}(6-2) = \log_{2}(8)-\log_{2}(4) \). This equals \( \log_{2}(2) = 1 \), confirming that the solution is correct.

Key Concepts

Properties of Logarithms

Logarithms have some unique and useful properties that make solving equations easier, especially when dealing with expressions involving differences or sums of logarithmic terms. One such property is that the difference of two logarithmic expressions with the same base can be rewritten as a single logarithm. This is because:

• For any two positive numbers, say \( A \) and \( B \), and a positive base \( b \) (not equal to 1), the property \( \log_{b}(A) - \log_{b}(B) = \log_{b} \left( \frac{A}{B} \right) \) holds true.

In practice, this means you can combine multiple logarithms into a single term, often simplifying calculations significantly. In our original exercise, using this property changed the equation from two separate log terms to a single one: \( \log_{2} \left( \frac{y+2}{y-2} \right) = 1 \). This simplification set the stage for further solving, ultimately making it more manageable to work through.

Exponential Form

The idea of converting logarithmic equations into their exponential form can transform an equation into something more familiar and easier to solve. This stems from the fundamental relationship between logarithms and exponents:

• The equation \( \log_{b}(x) = c \) is equivalent to \( x = b^c \).

By rewriting the logarithmic equation in this way, we convert a log expression into a straightforward exponential equation. In our problem, we transformed \( \log_{2} \left( \frac{y+2}{y-2} \right) = 1 \) into \( \frac{y+2}{y-2} = 2^1 \), simplifying it to \( \frac{y+2}{y-2} = 2 \). This type of conversion is particularly helpful because it often reduces the complexity of solving logarithmic equations, allowing us to apply our standard algebraic skills more effectively.

Checking Solutions

Verifying solutions is a vital step in solving mathematical problems as it ensures the accuracy of our results. After solving for \( y \) in our exercise, we found \( y = 6 \). But how do we know this is truly correct? We check by substituting it back into the original equation to see if it holds:

• The equation is \( \log_{2}(y+2) - \log_{2}(y-2) = 1 \).
• Substituting \( y = 6 \), this becomes \( \log_{2}(8) - \log_{2}(4) \).
• The result simplifies to \( \log_{2}(2) \), which indeed equals 1.

This alignment confirms that \( y = 6 \) is a valid solution. Checking solutions not only verifies correctness but also helps understand the behavior and constraints of our solutions within the original context of the problem.

Isolating Variables

Solving equations often involves isolating variables, a critical step in determining the value of unknowns. In our exercise, we had a fraction \( \frac{y+2}{y-2} = 2 \). To solve for \( y \), we needed to clear the fraction, simplifying the equation to a linear form:

• We first eliminated the fraction by multiplying both sides by \( y-2 \), resulting in \( y+2 = 2(y-2) \).
• Distributing and rearranging terms allowed us to gather \( y \) terms on one side: \( y + 2 = 2y - 4 \).
• Finally, isolating \( y \) yielded \( y = 6 \).

This process of isolating variables is invaluable because it turns complex expressions into solvable equations, using properties of equality and basic arithmetic to find our desired unknowns.