Problem 580 - Precalculus

Question

Use the one-to-one property of logarithms to find an exact solution for \(\log \left(4 x^{2}-10\right)+\log (3)=\log (51)\) If there is no solution, write no solution.

Step-by-Step Solution

Answer

The solutions are \( x = \pm \frac{3\sqrt{3}}{2} \).

Step 1: Simplify the Logarithmic Equation

Start by applying the Product Rule for logarithms to the left side of the equation. The Product Rule states that \( \log(a) + \log(b) = \log(a \cdot b) \), so we have:\[\log\left((4x^2 - 10) \cdot 3\right) = \log(51)\] This simplifies to:\[\log(12x^2 - 30) = \log(51)\]

Step 2: Apply the One-to-One Property

The one-to-one property of logarithmic functions states that if \( \log(a) = \log(b) \), then \( a = b \). Apply this property to the current equation:\[12x^2 - 30 = 51\]

Step 3: Solve the Quadratic Equation

To find the value of \( x \), solve the quadratic equation \( 12x^2 - 30 = 51 \):1. Add 30 to both sides: \[ 12x^2 = 81 \]2. Divide both sides by 12: \[ x^2 = \frac{81}{12} \] Simplify \( \frac{81}{12} \) to \( \frac{27}{4} \).3. Take the square root of both sides: \[ x = \pm \sqrt{\frac{27}{4}} \] Simplifying the square root results in: \[ x = \pm \frac{3\sqrt{3}}{2} \]

Step 4: Check for Extraneous Solutions

Check if the solutions satisfy the original domain conditions of the logarithmic expression.- Substituting \( x = \frac{3\sqrt{3}}{2} \) into \( 4x^2 - 10 \), calculate: \[ 4\left(\frac{3\sqrt{3}}{2}\right)^2 - 10 = 4\left(\frac{27}{4}\right) - 10 = 27 - 10 = 17 > 0 \]- Substituting \( x = -\frac{3\sqrt{3}}{2} \) into \( 4x^2 - 10 \) gives the same positive result because \( x^2 \) is the same.Thus, both values satisfy the domain of the logarithm.

Key Concepts

One-to-one Property

The one-to-one property is a fantastic rule that helps us solve equations involving logarithms. It works because logarithms are functions that grow steadily; each input value (or argument) maps to a unique output value. The essence of this property lies in simplicity: if you have two logarithmic expressions that are equal, like \( \log(a) = \log(b) \), you can confidently conclude that \( a = b \).

This property is particularly useful when solving equations, as it allows us to bypass the logarithm entirely and focus on the underlying numbers.

Whenever you see an equation involving a logarithm on both sides, think about using the one-to-one property to make your life easier — it transforms a potentially complex problem into a straightforward algebraic equation.

Product Rule

Understanding the Product Rule for logarithms can be a lifesaver when simplifying these expressions. It's like a secret weapon in your mathematical toolkit that transforms addition into multiplication. The rule states that \( \log(a) + \log(b) \) is equivalent to \( \log(a \cdot b) \).

This is incredibly handy because multiplying numbers together can sometimes be simpler than adding their logarithms separately.

When you see a sum of two logs, translating it using the product rule can streamline the process, especially when you're working to isolate variables or simplify complex equations. Apply this rule liberally when simplifying logarithmic equations before using other properties, like the one-to-one property.

Quadratic Equation Solution

Quadratic equations can seem daunting, but with a structured approach, they become manageable. In the exercise above, after applying properties of logarithms, you reach a quadratic equation: \( 12x^2 - 30 = 51 \).

Once you're at this point, solving it involves a few simple steps:

• First, get rid of any constants on one side, often done by adding or subtracting them from both sides.
• Next, isolate the quadratic term, for instance, by dividing through by any coefficients.
• Finally, take the square root of both sides, remembering to consider both the positive and negative roots.

In our case, this involved isolating \( x^2 \), then solving for \( x \) by taking the square root, leading to solutions \( x = \pm \frac{3\sqrt{3}}{2} \). Solving quadratic equations often ends with checking these solutions against original conditions to ensure they don't lead to undefined logarithmic expressions.