Problem 9 - Elementary and Intermediate Algebra

Question

Use a calculator to verify that equation is true. See Using Your Calculator: Verifying Properties of Logarithms. \(\ln \frac{11.3}{6.1}=\ln 11.3-\ln 6.1\)

Step-by-Step Solution

Answer

The equation is true; both sides equal approximately 0.619459.

Step 1: Calculate Right Side of Equation

First, we will calculate \( \ln 11.3 \) and \( \ln 6.1 \) separately using a calculator. \( \ln 11.3 \approx 2.427748 \) and \( \ln 6.1 \approx 1.808289 \). Then, compute \( \ln 11.3 - \ln 6.1 \). Using the values from the calculator, this is approximately \( 2.427748 - 1.808289 \approx 0.619459 \).

Step 2: Calculate Left Side of Equation

Now, compute \( \ln \left(\frac{11.3}{6.1}\right) \) using a calculator. First, find the division \( \frac{11.3}{6.1} \approx 1.852459 \). Then compute \( \ln 1.852459 \) which is approximately \( 0.619459 \).

Step 3: Verify the Equation

Compare the values calculated in Step 1 and Step 2. \( \ln 11.3 - \ln 6.1 \approx 0.619459 \) and \( \ln \left(\frac{11.3}{6.1}\right) \approx 0.619459 \), confirming both sides of the equation are equal. Hence, the logarithmic property that \( \ln \left(\frac{a}{b}\right) = \ln a - \ln b \) holds true in this case.

Key Concepts

Using a Calculator

Calculators are handy tools for solving mathematical problems efficiently. When working with logarithms, the calculator can be your best friend in determining complex logarithmic values. In our exercise, we used it to compute the natural logarithm, denoted as \( \ln \). The \( \ln \) function on calculators gives us the power to quickly find the logarithm of a number with base \( e \) (approximately 2.71828).

To solve an equation like \( \ln \frac{11.3}{6.1} = \ln 11.3 - \ln 6.1 \), you can start by pressing the 'ln' button on your calculator for each number to get their log values. Make sure to enter them precisely:

• Find \( \ln 11.3 \). This is around 2.427748.
• Next, find \( \ln 6.1 \). This returns approximately 1.808289.
• Subtract the two values to find \( \ln 11.3 - \ln 6.1 \).

Lastly, check if this result matches \( \ln \left(\frac{11.3}{6.1}\right) \) to confirm your equation's correctness. With these steps, calculators become indispensable in verifying complex mathematical identities.

Logarithmic Identities

Logarithmic identities are fundamental rules that simplify logarithmic expressions and solve equations. They describe the relationships between logarithms and their operands. For instance, one basic identity is \( \ln \left(\frac{a}{b}\right) = \ln a - \ln b \). This rule, often referred to as the Quotient Rule, lets us break down and understand logarithms of fractions easily.

So, why is it useful? Consider trying to find \( \ln \left(\frac{11.3}{6.1}\right) \) directly and without a calculator. By using the quotient identity, we simply find \( \ln 11.3 \) and \( \ln 6.1 \) separately, and subtract:

• Quotient Rule: \( \ln \left(\frac{a}{b}\right) = \ln a - \ln b \)

This makes dealing with bigger numbers less daunting and computational work more straightforward. Other identities like the Product Rule or Power Rule further enrich our mathematical toolkit, providing various avenues to solve and verify logarithmic equations.

Verifying Equations

Verifying equations is an essential process—as seen in the original problem—confirming that both sides of a mathematical expression are equal when calculated. In the context of logarithms, it involves checking if a logarithmic property holds true.

In our exercise, we verified \( \ln \frac{11.3}{6.1} = \ln 11.3 - \ln 6.1 \) using calculated values. Here's the responsible approach:

• First, compute the right-hand side (\( \ln 11.3 - \ln 6.1 \)).
• Then, compute the left-hand side (\( \ln \left(\frac{11.3}{6.1}\right) \)).
• Compare results: both should match for the identity to be verified.

This comparison helps us not only verify correctness but also enhances comprehension of why the identities serve as shortcuts in solving logarithmic equations. Establishing equation validity through this systematic approach fosters a deeper understanding of mathematical properties.