Problem 78 - Intermediate Algebra

Question

Solve each equation. $$ \log x+\log (x+9)=1 $$

Step-by-Step Solution

Answer

The solution is \( x = 1 \).

Step 1: Apply Logarithm Properties

Recall the logarithm property: \( \log a + \log b = \log(ab) \). Apply this to the equation to combine the logs: \[ \log x + \log(x+9) = \log(x(x+9)) = \log(x^2 + 9x). \] This gives us the equation: \( \log(x^2 + 9x) = 1 \).

Step 2: Convert to Exponential Form

Recall that if \( \log_b a = c \), then \( a = b^c \). Here, we are using base 10, so the equation \( \log(x^2 + 9x) = 1 \) converts to: \[ x^2 + 9x = 10^1 = 10. \] So, we have the quadratic equation: \( x^2 + 9x = 10 \).

Step 3: Rearrange into Standard Quadratic Form

Rearrange the equation to the standard quadratic form \( ax^2 + bx + c = 0 \) by subtracting 10 from both sides: \[ x^2 + 9x - 10 = 0. \] Now the equation is ready for solving by factoring or using the quadratic formula.

Step 4: Solve the Quadratic Equation by Factoring

Factor the quadratic equation \( x^2 + 9x - 10 = 0 \). We look for two numbers that multiply to \(-10\) and add to \(9\). These numbers are \(10\) and \(-1\). Thus, the factors are: \[ (x + 10)(x - 1) = 0. \]

Step 5: Find the Solutions

Set each factor equal to zero and solve for \(x\): \(x + 10 = 0\) or \(x - 1 = 0\). This gives the solutions \(x = -10\) and \(x = 1\).

Step 6: Verify Logarithm Validity

For a logarithm to be defined, the argument must be positive. Check the solutions: \(x = -10\) results in \(\log(-10)\), which is undefined. Thus, \(x = -10\) is not valid. For \(x = 1\), we have \(\log(1) + \log(1+9) = 0 + 1 = 1\), validating this solution.

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental concept in intermediate algebra. They are polynomial equations of degree 2, generally expressed in the form \( ax^2 + bx + c = 0 \). A quadratic equation can have two solutions, which we can find by various methods such as factoring, using the quadratic formula, or completing the square.

To solve a quadratic equation by factoring, the equation must be set to zero. The form \( ax^2 + bx + c = 0 \) is standard, and it allows us to find values of \( x \) that satisfy the equation. Solving by factoring involves finding two numbers that multiply to give the constant term \( c \) and add up to the coefficient of the linear term \( b \). This can be simple when the constants are integers.

• Example: Solving \( x^2 + 9x - 10 = 0 \).
• Look for two numbers that multiply to \(-10\) and add to \(9\).
• Factors found are \(10\) and \(-1\), forming \((x + 10)(x - 1) = 0\).

Set each factor to zero to solve for \( x \), resulting in potential solutions. However, you must verify each solution, especially when dealing with logarithmic bases.

Logarithmic Equations

Logarithmic equations involve logarithms and can initially appear daunting. They are equations of the form \( \log_b(x) = y \), which transforms into the exponential form \( x = b^y \). Logarithmic properties are crucial in simplifying and solving these equations.

In this exercise, using the property \( \log a + \log b = \log(ab) \), we simplify the given logarithmic expression. This property helps combine individual logarithms into a single logarithmic statement that can then be transformed into its exponential form.

• Given \( \log x + \log(x+9) = 1 \), apply the property to get \( \log(x(x+9)) = 1 \).
• This simplifies to \( \log(x^2 + 9x) = 1 \).
• Convert to exponential form: \( x^2 + 9x = 10^1 = 10 \).

Logarithmic equations often require converting back and forth between logarithmic and exponential forms, highlighting the deep connection between these concepts.

Factoring Polynomials

Factoring polynomials is a technique used to break down complex polynomial expressions into simpler, more manageable parts or factors. This process is essential in solving quadratic equations and can sometimes simplify them completely.

To factor a quadratic equation like \( x^2 + 9x - 10 \), search for two numbers that multiply to the constant term and add up to the coefficient of the linear term.

• Multiply: The factors of \(-10\) that add up to \(9\) are \(10\) and \(-1\).
• Breakdown: Express the quadratic as \((x + 10)(x - 1) = 0\).
• Solution Check: Set each factor equal to zero, \( x + 10 = 0 \) and \( x - 1 = 0 \).

These factors provide possible solutions to the quadratic equation, which you must then verify for validity in context, particularly for logarithmic considerations.

Exponential Equations

Exponential equations are equations where the variables appear as exponents. They typically take the form \( a^x = b \), with solutions often involving logarithmic expressions.

In the context of our exercise, we start with a logarithmic equation and convert it to exponential form to simplify for solving:

• From \( \log(x^2 + 9x) = 1 \), convert to exponential form: \( x^2 + 9x = 10 \).
• This involves the base 10 logarithm property \( b^y = x \).
• Now solve \( x^2 + 9x = 10 \) as a quadratic equation.

Understanding exponential equations requires a grasp on how they intertwine with logarithms, allowing one to switch back and forth between the two forms interchangeably, illustrating a key relationship in algebra.