Problem 70 - A Graphical Approach to Precalculus with Limits

Question

Use the properties of logarithms to rewrite each logarithm if possible. Assume that all variables represent positive real numbers. $$\log _{2} \frac{2 \sqrt{3}}{5 p}$$

Step-by-Step Solution

Answer

Rewrite as \( 1 + \frac{1}{2} \log_2 3 - \log_2 5 - \log_2 p \).

Step 1: Apply the Quotient Rule of Logarithms

The first step is to apply the quotient rule of logarithms which states \( \log_b \left( \frac{M}{N} \right) = \log_b M - \log_b N \). Here, \( M = 2\sqrt{3} \) and \( N = 5p \). We apply this rule as follows: \[ \log_2 \left( \frac{2\sqrt{3}}{5p} \right) = \log_2 (2\sqrt{3}) - \log_2 (5p) \].

Step 2: Apply the Product Rule of Logarithms

Apply the product rule to \( \log_2 (2\sqrt{3}) \), which states \( \log_b (MN) = \log_b M + \log_b N \). Here, \( M = 2 \) and \( N = \sqrt{3} \). Substitute to get: \[ \log_2 (2\sqrt{3}) = \log_2 2 + \log_2 \sqrt{3} \].

Step 3: Simplify the Logarithm of Powers

Next, simplify \( \log_2 \sqrt{3} \) and \( \log_2 (5p) \) by applying the power rule of logarithms, \( \log_b (M^n) = n \cdot \log_b M \). For \( \log_2 \sqrt{3} \), rewrite \( \sqrt{3} = 3^{1/2} \), thus: \[ \log_2 \sqrt{3} = \frac{1}{2} \cdot \log_2 3 \]. For \( \log_2 (5p) \), apply the product rule: \[ \log_2 (5p) = \log_2 5 + \log_2 p \].

Step 4: Combine the Logarithmic Expressions

Substitute the simplified expressions back into the equation from Step 1. You get: \[ \log_2 \left( \frac{2\sqrt{3}}{5p} \right) = (\log_2 2 + \frac{1}{2} \log_2 3) - (\log_2 5 + \log_2 p) \]. Simplify the expression further by using known logarithmic values, \( \log_2 2 = 1 \). This yields: \[ 1 + \frac{1}{2} \log_2 3 - \log_2 5 - \log_2 p \].

Key Concepts

Quotient Rule

The Quotient Rule is an essential property of logarithms that helps break down complex expressions into simpler terms. In simple words, it tells us that the logarithm of a division is equal to the difference between the logarithms of the numerator and the denominator. This rule can be a real timesaver for simplifying logarithmic expressions. Let's break it down with an example:

• Imagine you have an expression like \( \log_b \left( \frac{M}{N} \right) \).
• According to the Quotient Rule, this becomes \( \log_b M - \log_b N \).

In our exercise, we used the Quotient Rule to simplify \( \log_2 \left( \frac{2\sqrt{3}}{5p} \right) \) into \( \log_2 (2\sqrt{3}) - \log_2 (5p) \). This was the first step in making the original problem more manageable.
Breaking it down into two separate logs lets us handle each part individually with other rules, like the Product and Power Rules. The Quotient Rule makes working with complicated logarithmic expressions far simpler and clearer.

Product Rule

The Product Rule for logarithms allows us to break down the log of a product into a sum of logs. If you're looking at an expression where one term is a product, this rule can make your life easier. Here's how it works:

• For an expression \( \log_b (MN) \), the Product Rule states that this is equivalent to \( \log_b M + \log_b N \).

In our context, we had to simplify \( \log_2 (2\sqrt{3}) \). By applying the Product Rule:

• \( \log_2 (2\sqrt{3}) = \log_2 2 + \log_2 \sqrt{3} \).

Now, addressing each log individually becomes simpler. Each part of the product can be dealt with using other rules or known values. This approach highlights how the Product Rule makes complex logarithmic expressions much more manageable by distributing them into smaller parts.

Power Rule

The Power Rule is a handy tool when you have an exponent inside the log function. It allows you to "bring down" the exponent, turning it into a multiplier outside the log. Here's the gist:

• For any expression \( \log_b (M^n) \), the Power Rule tells us it's equivalent to \( n \cdot \log_b M \).

In the given exercise, we applied the Power Rule to expressions like \( \log_2 \sqrt{3} \). Since \( \sqrt{3} \) is \( 3^{1/2} \):

• Using the Power Rule, \( \log_2 \sqrt{3} = \frac{1}{2} \cdot \log_2 3 \).

This ability to pull down exponents in a log expression is powerful and often makes expressions much simpler to work with. The Power Rule is valuable when you're met with roots or higher-degree expressions inside logarithms, giving you a straightforward method to simplify the calculations at hand.