Problem 18 - Algebra 2 and Trigonometry

Question

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

Step-by-Step Solution

Answer

The value of \(x\) is approximately 3.65.

Step 1: Use Logarithmic Properties

To simplify the equation, use the property of logarithms that says \(a \cdot \log b = \log b^a\). Apply this to transform the left-hand side: \[2 \cdot \log x = \log x^2\].

Step 2: Simplify the Right-Hand Side

Apply the logarithmic property \(\log a + \log b = \log (a\cdot b)\) to the right-hand side:\[\log (x+3) + \log 2 = \log (2\cdot(x+3)) = \log (2x + 6).\]

Step 3: Set the Expressions Equal

Since the equation \log x^2 = \log (2x + 6)\ holds, we can set the arguments of the logarithms equal to each other:\[x^2 = 2x + 6.\]

Step 4: Form a Quadratic Equation

Rearrange the equation to form a standard quadratic equation \(x^2 - 2x - 6 = 0\).

Step 5: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1, b = -2, c = -6\):\[x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 24}}{2} = \frac{2 \pm \sqrt{28}}{2} = \frac{2 \pm 2\sqrt{7}}{2}.\]Simplifying gives \(x = 1 \pm \sqrt{7}\).

Step 6: Evaluate Numeric Values

Calculate the numeric solutions for \(x = 1 \pm \sqrt{7}\):\[x = 1 + \sqrt{7} \approx 3.65,\]\[x = 1 - \sqrt{7} \approx -1.65.\]

Step 7: Choose the Valid Solution

Because the logarithm of a negative number or zero is not defined, discard \(x = -1.65\). Thus, the valid solution is \(x = 3.65\).

Key Concepts

Quadratic Equations

Quadratic equations are expressions set equal to zero and are in the form of ax² + bx + c = 0, where a, b, and c are constants. In our exercise, after simplifying the logarithmic equation, we formed the quadratic equation x² - 2x - 6 = 0.
Quadratic equations can appear in various problem-solving contexts, including geometry, physics, and as in our case, integrating logarithmic expressions. Solving them provides important solutions, which we call roots.

• The roots can be real or complex numbers.
• The real roots are the x-intercepts of the equation when plotted on a graph.
• Knowing how to form quadratic equations is critical when dealing with more complex algebraic problems.

Once a quadratic equation is formulated, we can move on to finding the solutions using suitable methods, like factoring, completing the square, or employing the quadratic formula.

Logarithmic Properties

Logarithmic properties can simplify complex expressions and solve logarithmic equations more efficiently. In this particular exercise, we relied on two key properties:

• The Power Rule: This allows us to move a coefficient in front of a logarithm as an exponent of the argument. For example, 2 \(\log x = \log x^2\).
• The Product Rule: This enables the combination of two logarithmic expressions through multiplication. In our example, \(\log (x+3) + \log 2 = \log(2x + 6)\).

Understanding these properties helps us transform a logarithmic equation into a more manageable form. It is crucial when aiming to solve logarithmic equations, as it allows for isolating the variable of interest.

Quadratic Formula

The quadratic formula is a universal method used to find the roots of any quadratic equation and is expressed as:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Here, we take the coefficients a, b, and c from the equation to substitute them into the formula. For our equation x² - 2x - 6 = 0, the coefficients are a = 1, b = -2, and c = -6.
Applying the quadratic formula involves several steps:

• Find the discriminant \(b^2 - 4ac\).
• Determine if the discriminant is positive, negative, or zero to understand the nature of the roots.
• Calculate the roots using the \(\pm\) sign, which gives two potential solutions.

This formula is particularly advantageous because it does not require factoring and can handle any quadratic equation, even when it does not have rational roots.

Numeric Solutions

After using theoretical strategies to isolate and define the variable, obtaining numeric solutions involves evaluating expressions numerically. We calculated:

• \(x = 1 + \sqrt{7} \approx 3.65\)
• \(x = 1 - \sqrt{7} \approx -1.65\)

Although both results satisfy the quadratic equation, only \(x = 3.65\) makes sense in the context of our logarithmic equation.
Logarithms are only defined for positive numbers, meaning that solutions leading to negative evaluations must be disregarded. Therefore, when dealing with log-based equations, it is necessary to check the validity of the results to ensure that they fall within the allowed domain. This careful scrutiny guarantees that the final numeric solution is accurate and applicable.