Problem 11 - Algebra and Trigonometry Real Mathematics, Real People

Question

Determine whether each \(x\)-value is a solution of the equation. \(\log _{4}(3 x)=3\) (a) \(x \approx 21.3560\) (b) \(x=-4\) (c) \(x=\frac{64}{3}\)

Step-by-Step Solution

Answer

Among the provided x-values, \(x \approx 21.3560\) is a solution to the equation, whereas \(x=-4\) and \(x=\frac{64}{3}\) are not.

Step 1: Insert the first x-value

Substitute \(x \approx 21.3560\) into the equation to get \(\log _{4}(3*21.3560)\). Evaluate the right side of the equation to see if it equals to 3.

Step 2: Test whether x-value is a solution

After substitution and calculation, we find that \(\log _{4}(3*21.3560) \approx 3.0000. \approx 3\), therefore, \(x \approx 21.3560\) is a solution to the equation.

Step 3: Insert the second x-value

Substitute \(x=-4\) into the equation to get \(\log _{4}(3*-4)\). Evaluate the right side of the equation to see if it equals to 3.

Step 4: Test whether x-value is a solution

After substitution, we have an undefined log value because log function is undefined for negative values as the result would give us a complex number, so \(x=-4\) is not a solution to the equation.

Step 5: Insert the third x-value

Substitute \(x=\frac{64}{3}\) into the equation to get \(\log _{4}(3*\frac{64}{3})\). Evaluate the right side of the equation to see if it equals to 3.

Step 6: Test whether x-value is a solution

After substitution and calculation, we get \(\log _{4}(3* \frac{64}{3}) \neq 3\), thus \(x=\frac{64}{3}\) is not a solution to the equation.

Key Concepts

Logarithm Base Conversion

Understanding how to convert between bases of logarithms is essential for solving logarithmic equations. This task is often necessary when comparing logarithms with different bases or when simplifying expressions.

The formula for converting a logarithm from one base to another is: \ \[\log_{b}(x) = \frac{\log_{k}(x)}{\log_{k}(b)}\]

In this formula,\( b \) is the base we want to convert from, and \( k \) is the new base we are converting to. The key here is understanding that the logarithm of the number \( x \) with respect to base \( b \) can be expressed using any other base \( k \), which is typically base 10 or base e in calculations. Base 10 is the most common in calculators, known as the common logarithm denoted as \( \log \), and base \( e \) is the natural logarithm denoted as \( \ln \).

This process allows for flexibility and greatly facilitates complex calculations when solving logarithmic equations.

Solving Equations

Solving logarithmic equations involves finding the value(s) of the variable that make the equation true. There are several methods to tackle these types of equations, and one often needs to consider the properties of logarithms to simplify and solve them efficiently.

Here are some general steps to solve a logarithmic equation:

• Isolate the logarithmic expression, if possible.
• Convert the logarithmic equation into an exponential form. For instance, \( \log_{b}(x) = y \) implies \( b^{y} = x \).
• Solve the resulting equation for the variable.
• Check all potential solutions in the context of the original equation, because solutions must satisfy the condition that the argument of the logarithm is positive.

For example, in the equation \( \log _{4}(3x) = 3 \), convert to exponential form: \( 3x = 4^{3} \), leading to the solution \( x = \frac{64}{3} \).

Be aware that logarithmic functions have specific domains; thus, any proposed solutions resulting in undefined or negative arguments for the log function should be discarded.

Properties of Logarithms

Logarithms have several important properties that are crucial for simplifying and solving equations. A solid understanding of these properties allows for manipulation of expressions and solving equations more effectively.

Some fundamental properties include:

• \( \log_{b}(mn) = \log_{b}(m) + \log_{b}(n) \) - The product rule.
• \( \log_{b}(\frac{m}{n}) = \log_{b}(m) - \log_{b}(n) \) - The quotient rule.
• \( \log_{b}(m^{n}) = n \cdot \log_{b}(m) \) - The power rule.
• \( \log_{b}(b) = 1 \) because \( b^{1} = b \).
• \( \log_{b}(1) = 0 \) because \( b^{0} = 1 \).

By recognizing these properties, we can transform logarithmic expressions into simpler forms. This helps not only in solving equations but also in understanding their behavior.

In problems like the ones above, using these properties allows us to simplify the left-hand side of the equation, often making it more straightforward to solve. Remember, breaking down complex expressions using properties can significantly ease the solving process.