Problem 58 - Elementary and Intermediate Algebra

Question

Solve each equation. See Example \(9 .\) $$ \log 11-\log x=2 $$

Step-by-Step Solution

Answer

The solution is \( x = 0.11 \).

Step 1: Understand the Logarithmic Equation

The equation given is \( \log 11 - \log x = 2 \). The objective is to solve for \( x \). This is a logarithmic equation involving subtraction, which can be simplified using logarithmic properties.

Step 2: Use the Logarithmic Property

Utilize the logarithmic identity: \( \log a - \log b = \log \left( \frac{a}{b} \right) \). Apply this property to the equation: \( \log 11 - \log x = \log \left( \frac{11}{x} \right) \). Therefore, the equation becomes \( \log \left( \frac{11}{x} \right) = 2 \).

Step 3: Convert the Logarithmic Equation to Exponential Form

In order to solve the logarithmic equation, convert it into its exponential form. If \( \log_b A = C \), then \( b^C = A \). For this scenario, since the base is 10 (common logarithm), convert \( \log \left( \frac{11}{x} \right) = 2 \) to its exponential form: \( 10^2 = \frac{11}{x} \).

Step 4: Solve the Exponential Equation

Solve the equation \( 100 = \frac{11}{x} \) for \( x \). Cross-multiply to get \( 100x = 11 \). Then, divide both sides by 100 to isolate \( x \): \( x = \frac{11}{100} \).

Step 5: Simplify the Solution

Finally, express \( x \) in its simplest form. The solution is already simplified as \( x = \frac{11}{100} \), or equivalently, \( x = 0.11 \).

Key Concepts

Logarithmic Properties

Logarithmic properties are essential tools for simplifying complex logarithmic expressions. In logarithms, there are several important identities that we use to breakdown or combine terms:

• **Product Property**: \( \log_b(M \cdot N) = \log_b M + \log_b N \)
• **Quotient Property**: \( \log_b\left( \frac{M}{N} \right) = \log_b M - \log_b N \)
• **Power Property**: \( \log_b(M^p) = p \log_b M \)

Using these properties can transform equations, making them easier to solve. In our problem, the equation \( \log 11 - \log x = 2 \) uses the Quotient Property to combine the logs into a single expression. By doing so, we can then further manipulate the equation more easily using exponential forms.

Exponential Form

Converting logarithmic equations to exponential form is a powerful technique for solving them. The exponential form makes calculations more straightforward. In general, if you have \( \log_b A = C \), it means \( b^C = A \).
For our example, the equation after applying logarithmic properties is \( \log \left( \frac{11}{x} \right) = 2 \).
Given that this is a common logarithm (where the base is 10), the exponential form becomes \( 10^2 = \frac{11}{x} \). Now, it's much easier to handle as it allows us to solve for \( x \) using simple algebraic manipulations.

Solving for x

After converting our logarithmic equation into its exponential form, we solve it by isolating \( x \). From the expression \( 10^2 = \frac{11}{x} \), we know \( 10^2 = 100 \), hence the equation becomes \( 100 = \frac{11}{x} \).

Here's the step-by-step approach to solving it:

• Cross-multiply to eliminate the fraction: \( 100x = 11 \).
• Divide both sides by 100 to solve for \( x \): \( x = \frac{11}{100} \).

This fraction \( \frac{11}{100} \) simplifies directly to \( 0.11 \). By solving this, we've concluded the problem. These steps show how translating between different mathematical forms can simplify problem solving.