Problem 16 - Algebra 2 and Trigonometry

Question

Find \(x\) to the nearest hundredth. \(\log x+\log (x+2)=\log 3\)

Step-by-Step Solution

Answer

The value of \( x \) is \( 1.00 \).

Step 1: Apply Logarithm Property

Use the property of logarithms that states \( \log a + \log b = \log (ab) \). Apply this property to combine the logarithms: \( \log x + \log (x+2) = \log (x(x+2)) \).

Step 2: Set Equation Equal to Single Logarithm

The equation now becomes \( \log(x(x+2)) = \log 3 \). Since the logarithms are equal, set their arguments equal: \( x(x+2) = 3 \).

Step 3: Expand and Rearrange Equation

Expand the equation \( x(x+2) = 3 \) to \( x^2 + 2x = 3 \). Rearrange the equation to form a quadratic equation: \( x^2 + 2x - 3 = 0 \).

Step 4: Solve Quadratic Equation

Solve the quadratic equation \( x^2 + 2x - 3 = 0 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).In this equation, \( a = 1 \), \( b = 2 \), and \( c = -3 \).

Step 5: Calculate Using Quadratic Formula

Substitute the values into the quadratic formula: \( x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 12}}{2} = \frac{-2 \pm \sqrt{16}}{2} \).

Step 6: Simplify the Result

Simplify \( \frac{-2 \pm \sqrt{16}}{2} \) to find two potential solutions: \( x = \frac{-2 + 4}{2} = 1 \) or \( x = \frac{-2 - 4}{2} = -3 \). Since \( x \) must be positive in a logarithmic function, the solution is \( x = 1 \).

Step 7: Round to the Nearest Hundredth

The value of \( x = 1 \) is already rounded to the nearest hundredth. Thus, the final answer is \( x = 1.00 \).

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\). They are second-degree equations, meaning the highest exponent of the variable is 2. In our exercise, we have come across a quadratic equation after applying the properties of logarithms. This equation can be expressed as \(x^2 + 2x - 3 = 0\). Here, the coefficients are:

• \(a = 1\)
• \(b = 2\)
• \(c = -3\)

To find the solutions for \(x\), you can use different methods like factoring, completing the square, or the quadratic formula. Often, the quadratic formula is the most straightforward method, given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, the term \(b^2 - 4ac\) is called the discriminant, and it determines the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root.
• If it's negative, the equation has no real roots, only complex ones.

In our equation, substituting the values gives a discriminant of 16, which indicates two real roots. This understanding allows us to solve for \(x\) using precise algebraic steps.

Properties of Logarithms

Logarithms have properties that simplify expressions and equations involving logarithmic terms. These properties make it easier to manipulate logarithms, especially in equations. A key property used in this exercise is:

• Product Property: \(\log a + \log b = \log (ab)\)

In our exercise, we used this property to combine the terms \(\log x\) and \(\log (x+2)\), resulting in \(\log(x(x+2))\). This simplifies the equation by turning the sum of two logs into a single logarithm. Another important property is the Equality Property:

• If \(\log a = \log b\), then \(a = b\)

This property allowed us to equate the expressions inside the logs, transforming the logarithmic equation into a simpler algebraic form: \(x(x+2) = 3\). Understanding these properties is crucial for tackling logarithmic equations effectively, providing a clear path toward finding solutions.

Solving Equations

Solving equations involves finding the values of variables that satisfy the equation. When dealing with logarithmic equations, such as the one in our exercise, using properties of logarithms is essential in simplifying and solving them. Once simplified, the next step usually involves solving the resulting algebraic equation. In this case, after simplifying the logarithmic equation using the product property, we ended up with a quadratic equation.To solve this quadratic equation \(x^2 + 2x - 3 = 0\), we used the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula is particularly useful because it ensures that we consider all potential roots of the quadratic equation. By substituting \(a = 1\), \(b = 2\), and \(c = -3\) into the formula, we calculated two possible solutions for \(x\):

• \(x = 1\)
• \(x = -3\)

However, not all solutions obtained this way are valid in the context of the original logarithmic equation, as logs are undefined for non-positive values. Hence, the valid solution is \(x = 1\). This value fits within the domain of the original logarithmic expression, demonstrating the importance of checking if solutions meet all constraints of the problem.