Problem 10 - 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. $$ 5 \log \frac{2}{3} $$

Step-by-Step Solution

Answer

The logarithmic expression \(5 \log \frac{2}{3}\) converts to algebraic form as \(a^5 - b^5\) where \(\log 2=a\) and \(\log 3=b\).

Step 1: Exploring the Expression

Starting with \(5 \log \frac{2}{3}\), it is noticed that the logarithm is base 10 by convention since no other base is specified. Also, it involves division between 2 and 3. Here, quotient rule of logarithms can be applied which states: \(\log (M/N) = \log M - \log N\).

Step 2: Applying the Quotient Rule

Apply the quotient rule of logarithms in the expression. As a result, \(5 \log \frac{2}{3}\) becomes \(5 (\log 2 - \log 3)\). And after opening up the brackets, the expression can be rewritten as \(5 \log 2 - 5 \log 3\).

Step 3: Use of Power Rule

This step involves applying the power rule which states that: \(C \log M = \log M^C\). Hence the expression \(5 \log 2 - 5 \log 3\) can be written as \(\log 2^5 - \log 3^5\).

Step 4: Applying the Given Logarithms of 2 and 3

Now, substitute the given logarithms of 2 and 3 in place of \(\log 2\) and \(\log 3\) in the expression. Thus, \(\log 2^5 - \log 3^5\) becomes \(a^5 - b^5\).

Key Concepts

Logarithm Rules

Logarithms are mathematical keys that unlock exponents. They help us understand how many times a certain number, called the base, must be multiplied by itself to reach another number. To work with logarithms effectively, you need to understand some key rules.

• The Product Rule states that the logarithm of a product is the sum of the logarithms: \( \log(M \times N) = \log M + \log N \).
• The Quotient Rule states that the logarithm of a quotient is the difference of the logarithms: \( \log(\frac{M}{N}) = \log M - \log N \).
• The Power Rule is essential when you have exponents. It states that the log of a number raised to an exponent is the exponent times the logarithm of the base: \( C\log M = \log M^C \).

Understanding these rules allows you to manipulate logarithmic expressions and simplify complex calculations. Each rule is an invaluable tool in your mathematical toolkit, especially when dealing with exponential growth or decay scenarios.

Quotient Rule

The quotient rule of logarithms simplifies expressions where one term is divided by another in a logarithmic context. For example, if you have \( \log \frac{2}{3} \), instead of calculating it directly, you can apply the quotient rule. In our exercise, it was used simply as:- Begin with the expression: \( \log \frac{2}{3} \).- Decompose it using the rule: \( \log \frac{2}{3} = \log 2 - \log 3 \).Notice how it beautifully transforms the logarithm of a fraction into a difference of logarithms. This property is particularly useful when you have values for \( \log 2 \) and \( \log 3 \), allowing you to sidestep more complicated calculations. Remember: it’s like undoing the division that multiplication creates.

Power Rule

The power rule provides a neat way to tackle logarithms that involve exponents. When you encounter an expression like \( 5 \log M \), the power rule can simplify it significantly.Consider how it was applied in our exercise:- Start with \( 5 \log 2 - 5 \log 3 \).- Apply the power rule: \( 5 \log M = \log M^5 \). So the expression becomes \( \log 2^5 - \log 3^5 \).This step utilizes the power rule to raise the base (in these examples, 2 and 3) to the given power (5), simplifying the expression and making it easier to handle. Applying the power rule is akin to compressing a lengthy multiplication process into a single exponential function, saving you from cumbersome calculations when values are substituted later on.