Problem 86 - Intermediate Algebra

Question

For Problems \(81-97\), solve each of the equations. $$ \log _{3} x+\log _{3}(x-2)=1 $$

Step-by-Step Solution

Answer

The solution is \( x = 3 \).

Step 1: Understand the Logarithm Addition Property

Recall the property of logarithms: the sum of two logarithms with the same base can be expressed as the logarithm of the product of their arguments. Hence, \( \log_{3} x + \log_{3}(x-2) = \log_{3}(x(x-2)) \). This simplifies the expression to a single logarithm.

Step 2: Use the Logarithm Equation

Given the equation \( \log _{3}[x(x-2)] = 1 \), we know that if \( \log_{3} a = b \), then \( a = 3^{b} \). Apply this to obtain \( x(x-2) = 3^{1} = 3 \).

Step 3: Simplify the Quadratic Equation

From \( x(x-2) = 3 \), expand the equation to form a quadratic equation: \( x^2 - 2x = 3 \). Rearrange it as \( x^2 - 2x - 3 = 0 \).

Step 4: Solve the Quadratic Equation

To solve \( x^2 - 2x - 3 = 0 \), factor the quadratic equation. This can be factored as \( (x-3)(x+1) = 0 \).

Step 5: Find the Valid Solutions

Set each factor to zero: \( x - 3 = 0 \) gives \( x = 3 \), and \( x + 1 = 0 \) gives \( x = -1 \). Check which solutions are valid by substituting back into the original equation: since the logarithm argument must be positive, \( x = -1 \) is not valid. Thus, the only solution is \( x = 3 \).

Key Concepts

Properties of Logarithms

When working with logarithms, understanding their properties is key. One of the most widely used properties is the addition property. This property states that the sum of two logarithms with the same base is equivalent to the logarithm of the product of their arguments.

Formally, it's expressed as:

• \( \log_b (m) + \log_b (n) = \log_b (m \cdot n) \)

This transformation helps in simplifying and solving equations involving logarithms. In our equation, \( \log_3 x + \log_3 (x-2) = 1 \), we used this property to condense two separate logs into a single one: \( \log_3 (x(x-2)) \).

This kind of simplification can make it much easier to solve the equation since we are dealing with just one logarithmic expression on one side.Additionally, remember the definition of a logarithm: for \( \log_b (a) = c \), it means that \( b^c = a \). This becomes valuable when moving from logarithmic to exponential form, allowing us to solve for the variable inside the log.

Quadratic Equations

Once the logarithmic expression \( \log_3 (x(x-2)) = 1 \) is transformed using the properties of logarithms, it leads to an equation involving a quadratic expression. This is a very common development when working with logarithmic equations.

Quadratic equations are equations that can be written in the standard form:

• \( ax^2 + bx + c = 0 \)

In our case, transforming the logarithmic form into exponential form \( x(x-2) = 3 \) results in a quadratic equation \( x^2 - 2x - 3 = 0 \). Solving quadratic equations can involve several methods, including factoring, completing the square, or using the quadratic formula.

It’s essential to recognize the importance of arranging all terms on one side to make one side of the equation equal to zero, which is helpful for factoring or other solution techniques. Remember, finding the roots of a quadratic equation gives us possible solutions for the variable we're solving for.

Factoring

Factoring is a key step in solving quadratic equations, especially when the equation is easily divisible into simpler expressions.

In the equation \( x^2 - 2x - 3 = 0 \), recognizing patterns or applying techniques like trial and error can help uncover factors. Here, the quadratic factors into \((x - 3)(x + 1) = 0\).

Factoring transforms the original quadratic into two simpler linear expressions. From these expressions, setting each factor equal to zero gives potential solutions for \( x \):

• If \( x - 3 = 0 \), then \( x = 3 \)
• If \( x + 1 = 0 \), then \( x = -1 \)

Only the solution where \( x = 3 \) satisfies the condition for logarithms, which they require the argument to be positive. Thus, factoring not only aids in solving the equation but also in identifying valid solutions when constraints apply, such as those found in logarithmic functions.