Problem 12 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

In Problems 10 through 13, let \(\log 2=a\) and \(\log 3=b .\) Express each of the following in terms of a and \(b .\) There should be no logarithms explicitly in the expressions you give. $$ 5 \log \sqrt[3]{6} $$

Step-by-Step Solution

Answer

The expression \(5 \log \sqrt[3]{6}\) in terms of \(a\) and \(b\) is \(5/3 \cdot a + 5/3 \cdot b\).

Step 1: Simplify the logarithm

First, expression inside the logarithm can be expressed as \( \sqrt[3]{6} = 6^{1/3}\). So, we have \(5 \log (6^{1/3})\). Using the property of logs that takes a power out of the log, we can rewrite this to \(5 \cdot (1/3) \cdot \log 6\).

Step 2: Break down logarithm

Next, using the change of base formula, break down \(\log 6\) into \(a\) and \(b\). Since 6 can be expressed as 2*3, we get \(\log 6 = \log 2 + \log 3=a+b\). So, \(5 \cdot (1/3) \cdot \log 6\) is rewritten as \(5 \cdot (1/3) \cdot (a+b)\).

Step 3: Simplify the expression

Finally, simplify the expression by multiplying everything out. The final answer would thus be \(5/3 \cdot a + 5/3 \cdot b\).

Key Concepts

Change of Base Formula

The change of base formula is a pivotal concept when it comes to simplifying logarithmic expressions. It allows you to convert a logarithm from one base to a different base that might be more convenient for solving problems. The formula is expressed as
\[\begin{equation}\log_b(x) = \frac{\log_c(x)}{\log_c(b)}\end{equation}\]
where \( b \) and \( c \) represent different bases, and \( x \) is the argument of the logarithm. To see the change of base formula in action, consider the given exercise where \( \log 6 \) needs to be broken down in terms of \( a \) and \( b \), which are \( \log 2 \) and \( \log 3 \) respectively. Through the change of base formula, we can express \( \log 6 \) as \( \log 2 + \log 3 \) by understanding that 6 is the product of 2 and 3. This process simplifies complex logarithmic expressions, making them easier to manage and solve.

Logarithmic Expressions

Logarithmic expressions are the inverse operations of exponentiation. They are widely used to solve equations involving exponents as they allow us to unravel the powers and deal with the base numbers directly. For instance, the expression \( \log_b(x^a) \) can be simplified by using a logarithmic property that brings the exponent down:
\[\begin{equation}\log_b(x^a) = a \cdot \log_b(x)\end{equation}\]
This property was employed in step 1 of the solution, where the exponent \( \frac{1}{3} \) was brought in front of the logarithm. It is essential to understand these properties as they enable us to simplify logarithmic expressions to a more manageable form that involves basic arithmetic operations such as addition, subtraction, and multiplication.

Exponents and Logarithms

The relationship between exponents and logarithms is fundamental to understanding how to work with logarithmic equations. In essence, logarithms answer the question: to what exponent must we raise a base number to produce a given number. If you have an expression like \( b^x = y \), you can rewrite it in logarithmic form as \( \log_b(y) = x \).
Understanding this relationship clarifies how to deal with the step where the square root of a number, which is an exponent, gets converted into a logarithmic form. As demonstrated in the problem, when combining the properties of exponents with logarithms, it becomes straightforward to switch between the two, allowing for the simplification of seemingly complex expressions. Always keep in mind that the base of the logarithm should be consistent, and any change to the base should be handled via the change of base formula.