Problem 29 - Algebra A Combined Function

Question

Solve each equation. $$ \log _{2} x+\log _{2}(3 x+1)=1 $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{2}{3}\).

Step 1: Use the Properties of Logarithms

We'll begin by using the property of logarithms that states the sum of logarithms is the logarithm of the product: \[\log_b a + \log_b c = \log_b (a \cdot c)\]. Here, we can apply this to our equation: \[\log_2 x + \log_2 (3x+1) = \log_2 [x(3x+1)]=1\].

Step 2: Convert Logarithmic Equation to Exponential Form

The equation \(\log_2 [x(3x+1)] = 1\) can be rewritten in exponential form. Recall that if \(\log_b y = c\), then \(y = b^c\). Here we have: \[x(3x+1) = 2^1 = 2\].

Step 3: Set Up and Solve the Quadratic Equation

Now substitute and arrange the terms from the previous step to form a quadratic equation: \[3x^2 + x - 2 = 0\]. This can be solved using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 1\), and \(c = -2\).

Step 4: Calculate the Roots Using the Quadratic Formula

Substitute the values into the quadratic formula: \[x = \frac{-1 \pm \sqrt{(1)^2 - 4(3)(-2)}}{2(3)}\].\[x = \frac{-1 \pm \sqrt{1 + 24}}{6}\].\[x = \frac{-1 \pm \sqrt{25}}{6}\].\[x = \frac{-1 \pm 5}{6}\]. Thus, the possible solutions are: \[x = \frac{4}{6} = \frac{2}{3}\] and \[x = -1\].

Step 5: Check the Validity of the Solutions

Since logarithms are not defined for zero or negative numbers, we need to check the validity of each solution.\(-1\) is not valid since \(\log_2(-1)\) does not exist.Therefore, the only valid solution is \(x = \frac{2}{3}\).

Key Concepts

Properties of Logarithms

Understanding the properties of logarithms is crucial in solving logarithmic equations. Logarithms have several essential properties that can simplify and transform expressions. Here, we used the product property, which states that the sum of logarithms with the same base can be written as a single logarithm:

• Sum of logarithms: \(\log_b a + \log_b c = \log_b (a \cdot c)\)

This property allows us to combine multiple logarithms into one, making the equation more manageable. In our example, \(\log_2 x + \log_2 (3x+1)\) becomes \(\log_2[x(3x+1)]\). This transformation places us one step closer to solving the equation.
Using properties like this is a powerful technique to simplify expressions, especially when dealing with logarithms. They help students reduce complexity in problems.

Exponential Form

After simplifying the logarithmic expression, the next step is to switch from a logarithmic to an exponential form. This conversion is essential because exponential equations are often easier to solve. The basic idea is based on the property:

• If \(\log_b y = c\), then \(y = b^c\).

By applying this definition to \(\log_2 [x(3x+1)] = 1\), we translate it to \(x(3x+1) = 2^1 = 2\). Switching to exponential form effectively removes the logarithm from the equation and allows for an algebraic approach to find the variable.
This step underscores the flexibility between different mathematical expressions, showing how logarithmic equations can be transformed into a more familiar format.

Quadratic Equation

Once transformed into exponential form, the equation often becomes a type of polynomial equation, such as a quadratic equation. In our example, we get the equation \(3x^2 + x - 2 = 0\) after simplifying the terms.Quadratic equations typically follow the form \(ax^2 + bx + c = 0\) and can be solved using various methods, such as factoring, completing the square, or the quadratic formula. We chose the quadratic formula in this case:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Substituting \(a = 3\), \(b = 1\), and \(c = -2\) gives us potential solutions. Factoring is sometimes possible, but the quadratic formula ensures that roots can be found even when they are not easily factorable.

Validity of Solutions

Once possible solutions are obtained, it is vital to determine their validity, especially in logarithmic equations.Logarithmic functions are only defined for positive real numbers because logarithms of zero or negative numbers do not exist in the real number system. Thus, when solutions such as \(x = -1\) appeared, they had to be checked:

• Check whether the values allow the logarithm to remain defined.
• Exclude values where \(\log_b x\) would result in an undefined expression.

Here, although we arrived at \(x = \frac{2}{3}\) and \(x = -1\), only \(x = \frac{2}{3}\) is valid, as the other solution results in taking the logarithm of a negative number.
This consideration affirms the importance of interpreting mathematical constraints contextually, ensuring that mathematical solutions align with defined domains of functions.