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Chapter 11: Problem 24

Solve each equation. $$2 \cdot \log (x)+\log (3)=\log (2-5 x)$$

Short Answer

Expert verified
The solutions are \(x = \frac{1}{3}\) and \(x = -2\).

Step by step solution

01

- Combine Logarithmic Terms

Use the properties of logarithms to combine the terms on the left side of the equation. The equation is initially given as:\(2 \cdot \log (x) + \log (3) = \log (2 - 5x)\)Transform \(2 \cdot \log (x)\) using the power rule: \(2 \cdot \log (x) = \log (x^2)\).Now combine the logs: \(\log (x^2) + \log (3) = \log (2 - 5x)\).
02

- Use the Addition Property of Logarithms

Use the property that \(\log (a) + \log (b) = \log (ab)\) to combine the logarithms on the left side:\(\log (x^2 \cdot 3) = \log (2 - 5x)\).Simplify the expression:\(\log (3x^2) = \log (2 - 5x)\).
03

- Equate the Arguments of the Logarithms

Since the logarithms are equal, their arguments must also be equal. Therefore, set the inside expressions equal to each other:\(3x^2 = 2 - 5x\).
04

- Solve the Quadratic Equation

Rearrange the equation to form a standard quadratic equation:\(3x^2 + 5x - 2 = 0\).Use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 5\), and \(c = -2\), to solve for \(x\).Calculate the discriminant:\(b^2 - 4ac = 5^2 - 4 \cdot 3 \cdot (-2) = 25 + 24 = 49\).Now solve for \(x\):\[x = \frac{-5 \pm \sqrt{49}}{6}\]This simplifies to:\[x = \frac{-5 + 7}{6} = \frac{2}{6} = \frac{1}{3}\]and\[x = \frac{-5 - 7}{6} = \frac{-12}{6} = -2\].
05

- Check for Extraneous Solutions

Verify whether the solutions satisfy the original logarithmic equation. For each solution, check if it keeps the arguments of the logarithms positive:For \(x = \frac{1}{3}\):\(2 - 5x = 2 - 5 \cdot \frac{1}{3} = 2 - \frac{5}{3} = \frac{6}{3} - \frac{5}{3} = \frac{1}{3} > 0\)For \(x = -2\):\(2 - 5x = 2 - 5 \cdot (-2) = 2 + 10 = 12 > 0\).Both \(x = \frac{1}{3}\) and \(x = -2\) are valid solutions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Properties of Logarithms
Understanding the properties of logarithms is crucial when solving logarithmic equations. These properties help simplify complex logarithmic expressions.
Three key properties are used here:
  • Power Rule: \( a \, \log(b) = \log(b^a) \)
  • Addition Property: \( \log(a) + \log(b) = \log(ab) \)
  • Equality Property: When \( \log(a) = \log(b) \), then \( a = b \)
Let's see how these properties are applied:
Initially, we have \( 2 \cdot \log(x) + \log(3) \).
Using the Power Rule, \( 2 \cdot \log(x) \) becomes \( \log(x^2) \). Next, we apply the Addition Property which states that adding logarithms is equivalent to the logarithm of a product. Hence, \( \log(x^2) + \log(3) = \log(3x^2) \).
This simplification is essential to equate the arguments of the logarithms in the next steps.
Quadratic Equation
A quadratic equation has the standard form \( ax^2 + bx + c = 0 \). Solving for \( x \) involves finding the values that satisfy this equation.
In this problem, we transform the logarithmic equation into the quadratic form \( 3x^2 + 5x - 2 = 0 \). To solve this, we use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
Here, \( a = 3 \), \( b = 5 \), and \( c = -2 \).
We first calculate the discriminant \( b^2 - 4ac = 49 \).
Plugging this into the formula gives:
  • \( x = \frac{-5 + 7}{6} = \frac{1}{3} \)
  • \( x = \frac{-5 - 7}{6} = -2 \)
These solutions indicate potential values for \( x \) but need verification in the original context to check for extraneous solutions.
Logarithmic Functions
Logarithmic functions are the inverse of exponential functions, helping us solve for variables in an exponential relationship.
For instance, the function \( y = \log_b(x) \) means finding the power to which the base \( b \) must be raised to get \( x \).
They follow strict rules: the argument of a logarithm must be positive.
In the given equation, checks must be performed on solutions obtained to ensure they don't apply on negative arguments as logarithms of negative numbers or zero are undefined.
The concept of logarithmic functions helps in reducing complex relationships to simpler algebraic forms, making such equations easier to solve.
Extraneous Solutions
Extraneous solutions are solutions that emerge from the solving process but do not satisfy the original equation. They often occur in logarithmic and rational equations.
For accuracy:
  • Verify the solutions in the context of the logarithmic functions.
  • Only accept solutions that satisfy the constraint that the argument of every logarithm is positive.
In our problem, both \( x= \frac{1}{3} \) and \( x= -2 \) solutions are correct upon substituting back:
For \( x = \frac{1}{3} \), the argument of the log function remains positive \( 2 - 5(\frac{1}{3}) = \frac{1}{3} \).
For \( x = -2 \), the argument \( 2 - 5(-2) \) is \( 12 \) which is again positive.
Errors can arise without checking this and accepting inappropriate solutions could lead to incorrect conclusions.

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