Problem 48 - Precalculus

Question

Solve the logarithmic equations exactly. $$\log _{2}(x-1)+\log _{2}(x-3)=3$$

Step-by-Step Solution

Answer

The solution is \(x = 5\).

Step 1: Understand the Problem

We are given the equation \(\log _{2}(x-1)+\log _{2}(x-3)=3\) and asked to solve for \(x\). These are logarithms with base 2, and we know that the sum of logs is equivalent to the log of the product.

Step 2: Apply the Product Rule for Logarithms

The sum of two logarithms with the same base can be rewritten as a single logarithm: \(\log _{2}(x-1) + \log _{2}(x-3) = \log _{2}((x-1)(x-3))\). So, the equation becomes \(\log _{2}((x-1)(x-3)) = 3\).

Step 3: Convert Logarithm to Exponential Form

We know that if \(\log_b A = C\), then \(b^C = A\). Therefore, \(\log_2((x-1)(x-3)) = 3\) can be rewritten as \(2^3 = (x-1)(x-3)\).

Step 4: Simplify the Equation

Since \(2^3 = 8\), we have \((x-1)(x-3) = 8\). This is a quadratic equation and can be expanded to \(x^2 - 4x + 3 = 8\).

Step 5: Solve the Quadratic Equation

Rearrange the equation to \(x^2 - 4x + 3 - 8 = 0\) which simplifies to \(x^2 - 4x - 5 = 0\). Factor the quadratic equation to \((x-5)(x+1) = 0\).

Step 6: Find the Roots of the Equation

Set each factor equal to zero: \(x-5 = 0\) gives \(x = 5\), and \(x+1 = 0\) gives \(x = -1\).

Step 7: Verify the Solutions

Substitute back to check both solutions. For \(x = 5\), \(\log_2(5-1)\) and \(\log_2(5-3)\) evaluates, which satisfies the equation. For \(x = -1\), \(x-1 = -2\), which cannot be computed in logarithms. Hence, \(x = -1\) is not valid.

Key Concepts

Product Rule for Logarithms

When dealing with logarithms, one useful property to remember is the product rule. The product rule states that if you have two logs with the same base being added together, you can combine them into a single log. This is done by multiplying the insides of the logs. For instance, if you have

• \( \log_b A + \log_b B \)

This can be rewritten as:

• \( \log_b (A \times B) \)

In our exercise, the expression \( \log_2(x-1) + \log_2(x-3) \) became \( \log_2((x-1)(x-3)) \). By applying this rule, the problem becomes easier to handle, simplifying it significantly so you can directly solve for the unknown variable.

Quadratic Equations

Quadratic equations are a type of polynomial equation that typically look like \( ax^2 + bx + c = 0 \). These equations are characterized by their highest power of the variable being 2. In our exercise, we derived a quadratic equation when we expanded the product of \((x-1)(x-3)\). The expanded form was \

• \( x^2 - 4x + 3 = 8 \)

When solving a quadratic equation, one method is to factor it into two binomials, which can be set equal to zero. In this case, \( x^2 - 4x - 5 = 0 \) factored into \((x-5)(x+1) = 0\). By setting each factor equal to zero, we can find the solutions of the equation, also known as the roots.

Exponential Form Conversion

Converting a logarithmic equation into an exponential form is a key step in solving many logarithmic problems. The relationship between logarithms and exponents can be expressed as: if \( \log_b A = C \), then this is equivalent to

• \( b^C = A \)

Using this principle allows us to move from a logarithmic to an exponential form, enabling straightforward calculations. In our exercise, converting \( \log_2((x-1)(x-3)) = 3 \) to exponential form yielded \

• \( 2^3 = (x-1)(x-3) \)

This conversion simplifies the problem to managing basic arithmetic operations, revealing the path to the solution. Recognizing this transformation is crucial in expanding the range of equations you can solve with confidence.