Problem 54 - 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 Property

Start by using the property of logarithms that states \( ext{log}(a) + ext{log}(b) = ext{log}(ab)\). Apply this to the equation: \(\log x + \log (x+9) = \log(x(x+9)) = 1\).

Step 2: Change of Base to Exponential Form

Recognize that \(\log(a) = b\) means \(10^b = a\). Therefore, rewrite the equation \(\log(x(x+9)) = 1\) in exponential form: \(x(x+9) = 10^1\).

Step 3: Simplify the Equation

Now simplify the expression \(x(x+9)\): \(x^2 + 9x = 10\). This results in a quadratic equation: \(x^2 + 9x - 10 = 0\).

Step 4: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) for the equation \(x^2 + 9x - 10 = 0\), where \(a = 1\), \(b = 9\), and \(c = -10\).

Step 5: Calculate Discriminant

Calculate the discriminant: \(b^2 - 4ac = 9^2 - 4 \cdot 1 \cdot (-10) = 81 + 40 = 121\).

Step 6: Find Roots Using Quadratic Formula

Substitute into the quadratic formula: \(x = \frac{-9 \pm \sqrt{121}}{2}\). Simplifying gives \(x = \frac{-9 \pm 11}{2}\), resulting in two solutions: \(x = 1\) and \(x = -10\).

Step 7: Verify Logarithmic Constraints

Only positive values are valid in logarithmic expressions, so \(x = -10\) is invalid. Therefore, the solution is \(x = 1\).

Key Concepts

Logarithm Properties

The equation \( \log x + \log (x+9) = 1 \) requires the application of specific properties of logarithms.
One such property is the product rule for logarithms. It states that the sum of the logarithms of two numbers is the logarithm of their product: \( \log(a) + \log(b) = \log(ab) \).
In our exercise, by applying this property, we combine \( \log x \) and \( \log(x+9) \) into a single logarithm: \( \log(x(x+9)) \).

• This transforms the equation into one log term, simplifying further operations.
• Recognizing these properties can significantly streamline solving logarithmic equations.

Another essential principle in log equations is understanding the switch between logarithmic and exponential forms.
Remember, if \( \log(a) = b \), it implies \( 10^b = a \). This conversion helps in solving the equation when you have isolated the logarithmic term.

Quadratic Equations

Quadratic equations are crucial when dealing with transformations from logarithmic forms.In our problem, after combining logarithms, we convert the equation \( \log(x(x+9)) = 1 \) to its exponential form, \( x(x+9) = 10 \).
This simplifies to the quadratic equation \( x^2 + 9x - 10 = 0 \). Quadratic equations have a standard form: \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants.

• In our case, \( a = 1 \), \( b = 9 \), and \( c = -10 \).

These equations can often be solved by factoring, completing the square, or using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Recognizing and reformulating quadratic equations is a valuable skill that applies in various topics, including when handling problems stemming from logarithmic equations.

Discriminant

In quadratic equations, the discriminant is a powerful tool that determines the nature of the roots.The discriminant \( \Delta \) is calculated as \( b^2 - 4ac \). It is part of the quadratic formula, within the square root portion: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

• If \( \Delta > 0 \), there are two distinct real roots.
• If \( \Delta = 0 \), there is exactly one real root (or a repeated real root).
• If \( \Delta < 0 \), the roots are complex (no real roots).

In this exercise, calculating the discriminant for \( x^2 + 9x - 10 = 0 \) gives:
\( \Delta = 9^2 - 4 \cdot 1 \cdot (-10) = 121 \).
Since \( 121 \) is positive, we expect two distinct real roots.This confirms that the quadratic equation has two solutions, although one is invalid due to logarithmic constraints, as logarithms require positive arguments.