Problem 69 - 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. $$\log _{5} \frac{5 \sqrt{7}}{3 m}$$

Step-by-Step Solution

Answer

The expression simplifies to \( 1 + \frac{1}{2} \log_5 7 - \log_5 3 - \log_5 m \).

Step 1: Apply Logarithm of a Quotient

The logarithm of a quotient can be rewritten as the difference of two logarithms. Thus, apply the property \( \log_b \left( \frac{x}{y} \right) = \log_b x - \log_b y \). Here, we have: \[ \log_5 \left( \frac{5 \sqrt{7}}{3m} \right) = \log_5(5 \sqrt{7}) - \log_5(3m). \]

Step 2: Apply Logarithm of a Product in Both Terms

Now, apply the property \( \log_b(xy) = \log_b x + \log_b y \) to both terms: \[ \log_5 (5 \sqrt{7}) = \log_5 5 + \log_5 \sqrt{7} \] and \[ \log_5 (3m) = \log_5 3 + \log_5 m. \] Substitute these into the equation from Step 1: \[ \log_5 5 + \log_5 \sqrt{7} - (\log_5 3 + \log_5 m). \]

Step 3: Simplify Using Power and Simple Logarithm Properties

Recognize that \( \log_b b = 1 \) for any base \( b \), so \( \log_5 5 = 1 \). Additionally, apply the power rule \( \log_b (x^r) = r \log_b x \) to \( \log_5 \sqrt{7} \): \( \log_5 \sqrt{7} = \log_5 (7^{1/2}) = \frac{1}{2} \log_5 7 \). Substitute these back into the expression: \[ 1 + \frac{1}{2} \log_5 7 - \log_5 3 - \log_5 m. \]

Step 4: Arrange the Expression

Now simplify the expression into a more readable form: \[ 1 + \frac{1}{2} \log_5 7 - \log_5 3 - \log_5 m. \] Keep the terms that involve logarithms grouped together for clarity.

Key Concepts

Logarithm of a Quotient

The logarithm of a quotient is a fundamental property of logarithms that allows us to break down the logarithm of a division into the subtraction of two different logarithms. This property can be expressed as: \( \log_b \left( \frac{x}{y} \right) = \log_b x - \log_b y \).

Let's see how to apply it. Suppose you need to simplify \( \log_5 \left( \frac{5 \sqrt{7}}{3m} \right) \). By using the logarithm of a quotient property, this expression converts to \( \log_5(5 \sqrt{7}) - \log_5(3m) \).

This property is incredibly useful because it helps in breaking down more complex logarithmic expressions into simpler components, which are easier to manage individually. So, next time you encounter a quotient inside a log, remember that subtraction can make things simpler to handle!

Logarithm of a Product

Another crucial property of logarithms is the logarithm of a product. This property allows us to express the logarithm of a multiplication as the sum of the logarithms of the factors involved. This property is written as: \( \log_b(xy) = \log_b x + \log_b y \).

Consider the expression \( \log_5 (5 \sqrt{7}) \). By applying the product rule, it becomes \( \log_5 5 + \log_5 \sqrt{7} \). Similarly, for \( \log_5(3m) \), you get \( \log_5 3 + \log_5 m \).

Breaking a product into summable parts makes it much easier to subsequently apply other logarithmic properties or further simplifications. The power of this property lies in its ability to disentangle complex products into simpler actions that you can then focus on individually for easier computation.

Power Rule of Logarithms

The power rule of logarithms is another helpful property that comes in handy when dealing with exponents inside a logarithmic expression. This rule states that if you have an exponent within the logarithm, you can "pull down" the exponent as a multiplier. The rule is represented as: \( \log_b (x^r) = r \log_b x \).

Take the term \( \log_5 \sqrt{7} \), which can be rewritten as \( \log_5 (7^{1/2}) \). By applying the power rule, it transforms to \( \frac{1}{2} \log_5 7 \).

Utilizing this rule helps by simplifying expressions where roots or powers are involved. It sets the stage for clearer computations and is especially useful when combined with other properties, such as the ones we previously discussed, allowing for comprehensive simplification and manipulation of logarithmic expressions. Keep this rule in mind whenever you bump into powers within logs!