Problem 53 - Intermediate Algebra

Question

Solve each equation. $$ \log x+\log (x-48)=2 $$

Step-by-Step Solution

Answer

The solution to the equation is \(x=50\).

Step 1: Use Logarithmic Properties

Recall the property of logarithms: \(\log a + \log b = \log (ab)\). Apply this to the equation \(\log x + \log (x - 48) = \log (x(x - 48))\). Now the equation becomes: \(\log(x^2 - 48x) = 2\).

Step 2: Convert Logarithmic Equation to Exponential Form

Using the definition of a logarithm, \(\log_b a = c\) means \(b^c = a\). Here, with base 10, convert the equation: \(\log(x^2 - 48x) = 2\) becomes \(x^2 - 48x = 10^2\).

Step 3: Simplify the Equation

Simplify \(x^2 - 48x = 100\) by moving all terms to one side to form a quadratic equation: \(x^2 - 48x - 100 = 0\).

Step 4: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -48\), \(c = -100\).Calculate the discriminant: \(b^2 - 4ac = (-48)^2 - 4 \cdot 1 \cdot (-100) = 2304 + 400 = 2704\).Solve for \(x\):\[ x = \frac{48 \pm \sqrt{2704}}{2} \]\[ x = \frac{48 \pm 52}{2} \]

Step 5: Evaluate Solutions

Calculate both solutions: \(x = \frac{48 + 52}{2} = 50\) and \(x = \frac{48 - 52}{2} = -2\).Since \(x\) must be positive for the logarithm to be defined, only \(x = 50\) is valid.

Key Concepts

Properties of Logarithms

Logarithms have some very useful properties that make solving equations easier. One of these is the property that allows you to combine the sum of logarithms. Specifically, the property: \( \log a + \log b = \log (ab) \). This is known as the product property of logarithms. By using this property, we can take multiple logarithmic terms and combine them into a single term. It simplifies equations immensely, especially when dealing with equations like \( \log x + \log (x-48) = 2 \).

• Product Property: Helps simplify the sum of logs into one.
• Combines terms for easier manipulation of the equation.
• Useful to condense expressions.

Using logarithmic properties allows us to transform complex expressions into something more manageable, paving the way to solve the equation step by step.

Exponential Form

In solving logarithmic equations, converting them into exponential form is a crucial step. You often start with a logarithmic form such as \( \log_b a = c \). This states that \( b^c = a \). In this exercise, we handle base 10, which is often implied when the base isn't written out.

For example, consider \( \log(x^2 - 48x) = 2 \). To convert to exponential form, you express it as \( x^2 - 48x = 10^2 \). This conversion simplifies the equation into a familiar format that's easier to work with.

• Exponential Form: Simplifies the solving process by eliminating logs.
• Makes it easier to manipulate the equation algebraically.
• Provides a clear path towards forming quadratic or other polynomial equations.

Being comfortable with switching between log and exponential forms is essential, as it helps see the problem from a different angle, unveiling different solution pathways.

Quadratic Formula

Quadratic equations come up frequently when manipulating algebraic expressions. After simplifying and converting the expression into a quadratic format, like \( x^2 - 48x - 100 = 0 \), you can solve it using the quadratic formula. The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) provides a systematic approach to find the roots of any quadratic equation.

Understanding how to apply the quadratic formula involves:

• Identifying coefficients: Recognizing \( a, b, \text{ and } c \) in \( ax^2 + bx + c = 0 \).
• Calculating the discriminant \( b^2 - 4ac \): This determines the nature of the roots.
• Plugging values into the formula and solving: Results in two possible solutions.

In our example, using the quadratic formula to solve \( x^2 - 48x - 100 = 0 \) yields solutions \( x = 50 \) and \( x = -2 \). Since logarithms only accept positive arguments, we discard \( x = -2 \), leaving \( x = 50 \) as the valid solution.