Problem 10 - Elementary and Intermediate Algebra

Question

Use a calculator to verify that equation is true. See Using Your Calculator: Verifying Properties of Logarithms. \(\log \sqrt{24.3}=\frac{1}{2} \log 24.3\)

Step-by-Step Solution

Answer

The equation is true; both calculations give approximately 0.683.

Step 1: Understand the Property

We are going to verify the property of logarithms: \(\log \sqrt{a} = \frac{1}{2} \log a\). This is derived from the logarithm power rule \(\log a^b = b \log a\).

Step 2: Identify the Value

Given the problem \(\log \sqrt{24.3}\), we recognize this as \(\log (24.3^{1/2})\) since \(\sqrt{a}\) is the same as \(a^{1/2}\).

Step 3: Calculate \(\log \sqrt{24.3}\) with a Calculator

Using a calculator, input \(\sqrt{24.3}\) to compute its value, then take the logarithm of the result. The precise value for \(\log \sqrt{24.3}\) should be around 0.683. Remember that log base 10 is used unless stated otherwise.

Step 4: Calculate \(\frac{1}{2} \log 24.3\) with a Calculator

First, calculate \(\log 24.3\) using your calculator. This value should be around 1.366. Now divide this result by 2, yielding an approximate value of 0.683.

Step 5: Compare Values

The two values from Step 3 and Step 4 are compared. Both are approximately 0.683, verifying the property \(\log \sqrt{24.3}=\frac{1}{2}\log 24.3\) is correct.

Key Concepts

Logarithm Properties

Logarithms have unique properties that make complex calculations simpler. One such property is the logarithm power rule, which states that \( \log a^b = b \log a \). This means that when you take the logarithm of an exponent, you can bring the exponent in front as a coefficient. This feature is immensely useful in simplifying logarithmic expressions.

• The property \( \log \sqrt{a} = \frac{1}{2} \log a \) comes directly from this rule. Recognize that \( \sqrt{a} \) can be expressed as \( a^{1/2} \).
• The essence of this property is translating the expression into a form that's easier to compute, especially when dealing with roots.

Understanding these foundational properties makes manipulating and verifying logarithmic equations far more intuitive. Once you internalize the rules, the properties become powerful tools in solving equations and understanding the relationships in logarithmic expressions.

Calculator Usage in Algebra

In algebra, a calculator is a critical tool for performing precise computations, especially with complex numbers or when working with logarithms. When verifying logarithmic equations, your calculator helps ensure accuracy and saves time.

• To compute \( \log \sqrt{24.3} \), first calculate \( \sqrt{24.3} \) using the square root function. Then, use the log function to find the logarithm of the result.
• For \( \frac{1}{2} \log 24.3 \), directly calculate \( \log 24.3 \) using the log function, and then divide the result by 2.

Ensure your calculator is set to the correct base (usually base 10) unless specified otherwise. The precision offered by calculators is unparalleled, making it easier to verify equations or check against expected logarithmic properties effectively.

Verifying Logarithmic Equations

Verifying logarithmic equations is an essential skill that demonstrates your understanding of logarithm properties. Once you've computed the necessary values using a calculator, comparing the results is crucial.

• With values such as \( \log \sqrt{24.3} \) and \( \frac{1}{2} \log 24.3 \), you compare their numerical outputs.
• If the two values match or are very close (accounting for minor computational discrepancies), the equation is verified correctly.

This process solidifies your knowledge of relevant logarithmic properties and integrates calculator usage for precise computations. This methodical approach to verifying helps reinforce confidence in handling logarithmic equations and strengthens the foundation for more advanced algebraic concepts.