Problem 34 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Use the laws of logarithms to solve the equation. $$\log _{x} \frac{1}{16}=-2$$

Step-by-Step Solution

Answer

The short answer: Solving the given equation \(\log_{x} \frac{1}{16} = -2\) using the laws of logarithms, we find that the base x has two possible values: \(x = \pm 4\).

Step 1: Understand the equation

First, take a moment to examine the given equation: \(\log_{x} \frac{1}{16} = -2\). Our goal is to find the value of the base x.

Step 2: Apply the rules of logarithm

Use the property, \(a^b = c\) means that \(a = \log_{b} c\). With that in mind, we can rewrite our given equation as: \(x^{-2} = \frac{1}{16}\).

Step 3: Solve the exponential equation

Now that we have the exponential equation, \(x^{-2} = \frac{1}{16}\), we can solve for x. This equation can also be written as \(x^2 = 16\).

Step 4: Find the square root

To find the value of the base, we should take the square root of both sides of the equation: \(\sqrt{x^2} = \sqrt{16}\).

Step 5: Determine the value of x

After solving the above equation, we obtain that \(x = \pm 4\). This means that our final solution is two possible values for the base x, which are (positive) \(4\) and (negative) \(-4\).

Key Concepts

Laws of Logarithms

When working with logarithmic equations, the laws of logarithms are your valuable allies. These rules help simplify expressions and solve equations involving logarithms. The main laws include:

• Product Law: This states that \( \log_b (MN) = \log_b M + \log_b N \).
• Quotient Law: This says \( \log_b \left( \frac{M}{N} \right) = \log_b M - \log_b N \).
• Power Law: This law implies \( \log_b (M^n) = n\log_b M \).

In this exercise, understanding that \( \log_b a = c \) implies that \( b^c = a \) was key. This conversion step allows us to transition from a logarithmic equation to an exponential one, setting the stage for easier calculations. Remembering how to switch between forms using the laws will make solving logarithmic equations more straightforward.

Exponential Equations

Exponential equations are those in which variables appear as exponents. To solve these, we often convert logarithmic forms into exponential ones, allowing us to work with more straightforward algebraic methods.
In the exercise, the equation \( \log_{x} \frac{1}{16} = -2 \) was converted into its exponential form, \( x^{-2} = \frac{1}{16} \). This step simplifies the problem. From here, we recognized a relationship with powers, specifically that \( x^{-2} \) simplifies to \( \frac{1}{x^2} \), which helped convert the problem into a standard quadratic form, \( x^2 = 16 \).
Understanding exponential equations means recognizing these transformations. Once an equation is in exponential form, you can apply algebraic methods, such as taking roots or factoring, to find solutions.

Solving Equations

The task of solving equations involves finding the values that make the equation true. Here, it is essential to follow a strategic approach.
When faced with solving for \( x \) in \( x^2 = 16 \), taking the square root is a natural step. This gives \( \sqrt{x^2} = \sqrt{16} \), leading to \( x = 4 \) or \( x = -4 \). Consider both positive and negative roots due to the square root operation, resulting in two potential answers.
Solving logarithmic and exponential equations often requires:

• Rewriting the equation for clarity and simplicity.
• Using known algebraic methods to isolate the variable.
• Checking solutions to ensure they satisfy the original equation.

With practice, solving equations becomes an intuitive process. Recognizing patterns and leveraging mathematical properties will enhance your ability to tackle diverse problems.