Problem 5 - Basic Engineering Mathematics

Question

Write \(\log \left(\frac{8 \times \sqrt[4]{5}}{81}\right)\) in terms of \(\log 2, \log 3\) and \(\log 5\) to any base.

Step-by-Step Solution

Answer

\( 3 \log 2 + \frac{1}{4} \log 5 - 4 \log 3 \)

Step 1: Express the expression as a single logarithm

Start with the given expression \( \log \left( \frac{8 \times \sqrt[4]{5}}{81} \right) \) and recognize it involves the quotient and product of numbers within the logarithm. Apply the logarithmic identity \( \log \left( \frac{a}{b} \right) = \log a - \log b \). This allows us to write: \( \log \left( 8 \times \sqrt[4]{5} \right) - \log 81 \).

Step 2: Apply the product rule for logarithms

Use the log property \( \log(ab) = \log a + \log b \) to expand the first term: \( \log (8 \times \sqrt[4]{5}) \). This becomes \( \log 8 + \log \sqrt[4]{5} \).

Step 3: Simplify the logarithm of 8

Write 8 as \( 2^3 \) to express it using logarithms of prime factors. Thus, \( \log 8 = \log (2^3) = 3 \log 2 \).

Step 4: Simplify the logarithm of the fourth root of 5

Express \( \sqrt[4]{5} \) as a power of 5: \( \sqrt[4]{5} = 5^{1/4} \). Therefore, \( \log \sqrt[4]{5} = \log (5^{1/4}) = \frac{1}{4} \log 5 \).

Step 5: Simplify the logarithm of 81

Recognize that 81 is \( 3^4 \). Thus, \( \log 81 = \log (3^4) = 4 \log 3 \).

Step 6: Combine the simplified terms

Replace the expressions calculated in previous steps back into \( \log \left( 8 \times \sqrt[4]{5} \right) - \log 81 \). This will be: \((3 \log 2 + \frac{1}{4} \log 5) - 4 \log 3 \).

Step 7: Final expression in terms of the base logarithms

Carefully combine all terms obtained: \( 3 \log 2 + \frac{1}{4} \log 5 - 4 \log 3 \). This is the expression for the original problem in terms of \( \log 2, \log 3, \) and \( \log 5 \).

Key Concepts

Logarithmic Identities

Logarithmic identities are a set of rules that make handling logarithms much easier, especially when simplifying complex expressions. One of the most fundamental identities is the quotient rule, which states that:

• \( \log \left( \frac{a}{b} \right) = \log a - \log b \)

This rule is particularly useful when working with expressions that involve division within the logarithm.
Another important identity is the power rule, where the logarithm of a number raised to an exponent can be simplified:

• \( \log(x^n) = n \log x \)

These identities help break down and convert complex arguments of a logarithm into simpler terms, as seen in the example where we transformed \( \log \left( \frac{8 \times \sqrt[4]{5}}{81} \right) \) into separate terms involving \( \log 2, \log 3, \text{ and } \log 5 \). They are crucial tools in algebra for making log expressions more manageable.

Product Rule for Logarithms

The product rule for logarithms is another pivotal identity, simplifying the logarithm of a product into a sum of individual logarithms:

• \( \log(ab) = \log a + \log b \)

This rule converts a single logarithm involving a product into two separate logarithms. It is particularly useful when dealing with products inside a logarithmic expression.
For instance, in the given exercise, we used the product rule to expand \( \log(8 \times \sqrt[4]{5}) \) into \( \log 8 + \log \sqrt[4]{5} \). By applying this rule, the expression becomes more straightforward, allowing us to utilize prime factorization and other properties to further simplify. The product rule is a powerful tool, making complex logarithmic expressions more accessible by breaking them down into simpler parts.

Prime Factorization

Prime factorization is a method of expressing a number as a product of its prime factors. Using prime factorization alongside logarithms allows us to apply logarithmic identities more effectively.
By breaking down numbers into prime factors, we can express any number in terms of its powers, such as \( 8 = 2^3 \) or \( 81 = 3^4 \). When these expressions are put into logarithmic form, they become easier to manipulate.
Utilizing prime factorization enables you to simplify the expression \( \log 8 \) to \( 3 \log 2 \) and \( \log 81 \) to \( 4 \log 3 \). This conversion to prime factors allows you to use the power identity efficiently, transforming complex values into sums of simpler logarithms related to primes. Correctly applying prime factorization significantly aids in simplifying and understanding problems involving logs.

Exponent and Logarithm Relationship

The relationship between exponents and logarithms forms the backbone of understanding logs. A logarithm essentially answers the question: "To what power must a base be raised to produce a given number?"

• For example, \( \log_b(x) = y \) means \( b^y = x \).

In this sense, logarithms are the inverses of exponentiation. Understanding this relationship is key to solving expressions involving both exponents and logs.
When dealing with the fourth root of a number, such as \( \sqrt[4]{5} \), it's expressed using exponents as \( 5^{1/4} \). Applying the power rule, \( \log(5^{1/4}) \) simplifies to \( \frac{1}{4} \log 5 \), highlighting how exponentiation within logs simplifies to multiplication.
Grasping this concept helps tremendously when interpreting and transforming logarithmic expressions, as it connects two foundational mathematical operations: multiplication and addition (for exponents and logs) in a clear, understandable manner.