Problem 38 - Precalculus Mathematics for Calculus

Question

Use a calculator to evaluate the expression, correct to four decimal places. (a) \(\log 50\) (b) \(\log \sqrt{2}\) (c) \(\log (3 \sqrt{2})\)

Step-by-Step Solution

Answer

(a) 1.69897; (b) 0.1505; (c) 0.6276

Step 1: Evaluating \(\log 50\)

Use a calculator to find the logarithm of 50 to four decimal places. Enter 50 and press the logarithm (log) button. The calculator will display the result.\[\log 50 \approx 1.69897\]

Step 2: Evaluating \(\log \sqrt{2}\)

First, recognize that \(\sqrt{2}\) is equal to \(2^{0.5}\). Therefore, \(\log \sqrt{2} = \log (2^{0.5})\). Use the logarithmic property of powers, i.e., \(\log (a^b) = b \cdot \log a\).\[\log (2^{0.5}) = 0.5 \times \log 2\]Use the calculator to find \(\log 2\) and then compute the expression.Previously computed: \(\log 2 \approx 0.3010\).\[\log \sqrt{2} \approx 0.5 \times 0.3010 = 0.1505\]

Step 3: Evaluating \(\log (3 \sqrt{2})\)

Utilize the logarithm property \(\log (a \cdot b) = \log a + \log b\). Here, we need to calculate \(\log 3\) and \(\log \sqrt{2}\), then sum them.Previously known: \(\log 3 \approx 0.4771\) and \(\log \sqrt{2} \approx 0.1505\).\[\log (3 \sqrt{2}) = \log 3 + \log \sqrt{2}\]Add the values:\[\log (3 \sqrt{2}) \approx 0.4771 + 0.1505 = 0.6276\]

Key Concepts

Logarithm Properties

Logarithms might sound intimidating, but they are a powerful mathematical tool that helps us simplify complex calculations. They are used to solve exponential equations and can be understood as the inverse of exponentiation. Two key properties to remember are:

• **Power Rule**: The logarithm of a power, such as \(\log (a^b)\), can be simplified using this rule: \(b \cdot \log a\). Essentially, you pull the exponent out front as a multiplier.
• **Product Rule**: The logarithm of a product, like \(\log (a \cdot b)\), can be expressed as the sum of individual logarithms: \(\log a + \log b\).

Understanding these properties allows you to break down and simplify logarithmic expressions efficiently. By mastering these rules, calculating expressions such as \(\log \sqrt{2}\) or \(\log (3 \sqrt{2})\) becomes much easier. The power rule assisted in transforming the square root expression into a multiplication problem, while the product rule helped separate the compound expression effectively, simplifying your calculation process.

Evaluating Logarithms

Evaluating logarithms involves finding an equivalent expression or numerical value for a given logarithmic statement. This is where the logarithmic properties come in handy. Let's break it down using the provided step-by-step solutions:

• For \(\log 50\), you directly use the calculator's log function to find its numeric value.
• For \(\log \sqrt{2}\): Recognize that \(\sqrt{2}\) is the same as the exponent \(2^{0.5}\). Use the power rule to simplify the calculation to \(0.5 \times \log 2\). Then, use a calculator to determine \(\log 2\) and multiply the result by 0.5.

Finally, for \(\log (3 \sqrt{2})\), the problem is broken down using both the power and product rules. You first calculate the individual components \(\log 3\) and \(\log \sqrt{2}\), then sum them up to achieve the final answer. Working through these steps with known values simplifies the logarithmic evaluations considerably.

Calculator Usage in Mathematics

Modern calculators can perform many complex mathematical functions, including logarithms, with ease. Learning to use a calculator to evaluate logarithms will save you time and reduce chances of error.

• Most scientific calculators have a dedicated "log" button. To use it for base 10 logarithms, just enter the number you wish to evaluate, and press the "log" button.
• For logarithms of expressions, where properties like the power or product rule are involved, you should perform these adjustments first, and then use the calculator to find the results.

Moreover, calculators provide the accuracy needed for most academic and real-world applications, like our exercises requiring results to four decimal places. With practice, calculator usage becomes an invaluable skill for quick and precise computations in mathematics.