Problem 52 - A Graphical Approach to Precalculus with Limits

Question

Solve each logarithmic equation. Express all solutions in exact form. Support your solutions by using a calculator. $$\log x+\log (3 x-13)=1$$

Step-by-Step Solution

Answer

The exact solution is \( x = 5 \).

Step 1: Apply Logarithmic Identity

We start with the equation \( \log x + \log (3x-13) = 1 \). We know from the properties of logarithms that \( \log a + \log b = \log(ab) \). Applying this identity, the equation becomes \( \log(x(3x-13)) = 1 \).

Step 2: Rewrite the Equation in Exponential Form

The equation \( \log(x(3x-13)) = 1 \) can be rewritten in exponential form. Recall that \( \log_b(a) = c \) is equivalent to \( a = b^c \). Since the base is 10 (common logarithm), we rewrite this as \( x(3x-13) = 10^1 \) or \( x(3x-13) = 10 \).

Step 3: Expand and Solve the Quadratic Equation

After expanding \( x(3x-13) = 10 \), we get \( 3x^2 - 13x - 10 = 0 \). We will solve this quadratic equation using the quadratic formula, where \( a = 3 \), \( b = -13 \), and \( c = -10 \). The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Step 4: Calculate the Discriminant

First, calculate the discriminant \( b^2 - 4ac \). Here, \( b^2 = (-13)^2 = 169 \) and \( 4ac = 4 \times 3 \times (-10) = -120 \). Thus, the discriminant is \( 169 + 120 = 289 \).

Step 5: Solve for x Using Quadratic Formula

With the discriminant calculated, substitute it into the quadratic formula: \( x = \frac{13 \pm \sqrt{289}}{6} \). Since \( \sqrt{289} = 17 \), the solutions are \( x = \frac{13 + 17}{6} = 5 \) and \( x = \frac{13 - 17}{6} = -\frac{2}{3} \).

Step 6: Verify and Conclude the Solution

We verify solutions by substituting back into the original logarithmic equation. The value \( x = -\frac{2}{3} \) is not valid, as logarithms of negative numbers are undefined. Hence, \( x = 5 \) is the only valid solution.

Key Concepts

Properties of Logarithms

Logarithms have specific properties that make complex calculations more manageable. One such property is the product property, which states that the logarithm of a product is equal to the sum of the logarithms. For example, for any positive numbers \(a\) and \(b\), and any logarithm base \(c\):

• \( \log_c(a \cdot b) = \log_c a + \log_c b \)

This property is particularly useful when solving equations involving multiple logarithmic terms, as it allows us to combine these terms into a single logarithm, simplifying the equation considerably.
By applying this property to our example, the equation \( \log x + \log (3x-13) = 1 \) was transformed into \( \log (x(3x-13)) = 1 \). Understanding how to manipulate logarithms through their properties is essential for solving logarithmic equations efficiently.

Exponential Form

In mathematics, the connection between logarithms and exponents is profound. When solving logarithmic equations, rewriting them in exponential form can be a critical step. The foundational idea is based on the relationship: if \(\log_b(a) = c\), this is equivalent to saying \(a = b^c\). This transformation allows us to work with the numbers more directly, especially when logarithms are involved.
In our original problem, after applying the property of logarithms, we had \(\log(x(3x-13)) = 1\). Since this uses the common logarithm (base 10), we convert it to exponential form: \(x(3x-13) = 10^1\), or simply \(x(3x-13) = 10\).
Working in exponential form often makes it easier to handle equations, especially when it involves finding intersection points or solutions with real numbers. Transitioning between logarithmic and exponential forms is a powerful tool in mathematical problem-solving.

Quadratic Formula

The quadratic formula is a reliable method for finding the roots of a quadratic equation. A standard quadratic equation is expressed as \(ax^2 + bx + c = 0\). The solutions for \(x\) are given by the formula:

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

This formula is especially useful when factoring is difficult or impossible. It utilizes the discriminant \(b^2 - 4ac\), which determines the nature of the roots—real and distinct, real and repeated, or complex.
In our exercise, we derived the equation \(3x^2 - 13x - 10 = 0\). Plugging into the quadratic formula with \(a = 3\), \(b = -13\), and \(c = -10\) yields two potential solutions. Calculations show the roots as \(x = 5\) and \(x = -\frac{2}{3}\).
Always remember to verify the suitability of solutions in the context of the original problem. Here, \(x = -\frac{2}{3}\) was extraneous since it isn't valid in the logarithmic domain, leaving \(x = 5\) as the sole valid answer.