Problem 12 - College Algebra with Modeling and Visualization

Question

Find the exact value of each expression. (a) \(\log \sqrt{100,000}\) (b) \(\log \sqrt[3]{100}\) (c) \(2 \log \sqrt{0.1}\) (d) \(10 \log \sqrt[3]{10}\)

Step-by-Step Solution

Answer

\(\frac{5}{2}, \frac{2}{3}, -1, \frac{10}{3}\).

Step 1: Simplify the square root of 100,000

First, we recognize that \(100,000\) is the same as \(10^5\). The square root of a number can be expressed as raising the number to the power of \(\frac{1}{2}\). So, \(\sqrt{100,000} = (10^5)^{\frac{1}{2}} = 10^{5/2}\).

Step 2: Calculate the logarithm of 10 raised to a power

Using the logarithm rule \(\log(a^b) = b\log(a)\), we find \(\log(10^{5/2}) = \frac{5}{2}\log(10)\). Since \(\log(10) = 1\), the expression simplifies to \(\frac{5}{2}\).

Step 3: Simplify the cube root of 100

The cube root of a number \(n\) is \(n^{\frac{1}{3}}\). For \(100\), express it as \(10^2\), so \(\sqrt[3]{100} = (10^2)^{\frac{1}{3}} = 10^{2/3}\).

Step 4: Calculate the logarithm of 10 raised to another power

Using \(\log(a^b) = b\log(a)\), we have \(\log(10^{2/3}) = \frac{2}{3}\log(10)\). Since \(\log(10) = 1\), the result simplifies to \(\frac{2}{3}\).

Step 5: Simplify the square root of 0.1

First convert 0.1 to a fraction: \(0.1 = \frac{1}{10}\). The square root of 0.1 is \((\frac{1}{10})^{\frac{1}{2}} = 10^{-1/2}\).

Step 6: Calculate the logarithm for the square root of 0.1

Using the property \(\log(a^b) = b\log(a)\), \(\log(10^{-1/2}) = -\frac{1}{2}\log(10)\). Therefore, \(\log(0.1) = -\frac{1}{2}\).

Step 7: Multiply the logarithm by 2

The expression is \(2 \log \sqrt{0.1} = 2(-\frac{1}{2}) = -1\).

Step 8: Simplify the cube root of 10

The cube root is \(10^{1/3}\).

Step 9: Calculate the logarithm of the cube root of 10 and multiply by 10

Using \(\log(a^b) = b\log(a)\), we have \(\log(10^{1/3}) = \frac{1}{3}\log(10)\). Multiply this by 10: \(10 \cdot \frac{1}{3} = \frac{10}{3}\).

Key Concepts

Logarithm Rules

Logarithms are fascinating mathematical tools that help us deal with exponentiation in a more manageable way. The most basic rule to know is the power rule:

• If you have a logarithm of a power like \( \log(a^b) \), you can bring the exponent \( b \) in front, changing it to \( b \cdot \log(a) \).

This rule simplifies computations and is extensively used in solving problems involving logarithms. For instance, in the original exercise, you see this rule applied when calculating \( \log(10^{5/2}) \). By bringing down the exponent \( \frac{5}{2} \), the problem simplifies significantly. Another basic rule to remember is:

• The logarithm of the base itself is 1, i.e., \( \log(10) = 1 \).

Using these basic rules simplifies the process of working with logarithms, allowing us to express and solve expressions more easily.

Simplifying Radicals

Radicals are expressions that involve roots, such as square roots or cube roots. Simplifying them involves re-expressing them as powers.

• A square root, written as \( \sqrt{n} \), can be rewritten as \( n^{1/2} \).
• Similarly, a cube root, \( \sqrt[3]{n} \), can be expressed as \( n^{1/3} \).

This conversion is particularly useful when dealing with logarithms because it fits well into the power rule of logarithms. For example, simplifying \( \sqrt{100,000} \) to \( 10^{5/2} \) prepares it nicely for applying the logarithm power rule. Avoid the temptation to leave radicals in their initial form when logarithms are involved; rewriting them using rational exponents is the way to go.

Power of Ten

Understanding powers of ten is crucial, especially since the base of common logarithms is ten. Each whole number exponent indicates how many times to multiply ten by itself, and negative exponents represent division. For example:

• \( 10^5 \) is \( 100,000 \).
• \( 10^{-1} \) is \( 0.1 \), as seen in the exercise.

When expressions involve powers of ten, converting other numbers to base ten simplifies the process of logarithmic calculations. Moving across powers of ten just involves shifting the decimal point without direct calculation, which is why it's so useful in problems like these.
The uniqueness of base ten logarithms lies in their simplicity when computing logarithms for neat base ten numbers.

Exact Values in Logarithms

Finding exact values means breaking down the problem into simple calculations, often reducing to known or easily computable base ten values. When given expressions like \( \log \sqrt{0.1} \) and needing precise outcomes, pattern recognition can be key.

• Notice how \( \log(10) \) simplifies to 1, which is fundamental for solving more complex problems.
• Using known conversions, the derived exact values, such as \( \frac{5}{2}, \frac{2}{3}, -1, \) and \( \frac{10}{3} \), show how manageable otherwise tricky logarithmic expressions can become.

The goal is to transform these expressions into forms that allow easy mental checks or use of simple rules, leveraging the properties of logarithms effectively to achieve exactness.