Problem 26 - College Algebra

Question

In Exercises \(1-40,\) use properties of logarithms to expand each logarithmic expression as much as possible. Where possible, evaluate logarithmic expressions without using a calculator. $$ \log _{8}\left(\frac{64}{\sqrt{x+1}}\right) $$

Step-by-Step Solution

Answer

After expanding the logarithmic expression as much as possible, the expression becomes \( 2 - \frac{1}{2} \log _{8}(x+1) \).

Step 1: Apply the Quotient Rule

According to the quotient rule, the logarithm of a quotient is equal to the difference of the logarithms of the numerator and the denominator. Therefore, the expression can be expanded as: \[\log _{8}(64) - \log _{8}(\sqrt{x+1})\]

Step 2: Evaluate the Logarithm of 64 base 8

Recall that \( \log _{a}(a^p) = p \). Therefore, \( \log _{8}(8^2) = 2 \), since 64 is \( 8^2 \). So, the expression now is: \[2 - \log _{8}(\sqrt{x+1})\]

Step 3: Apply the Power Rule

For the square root term, convert it as an exponent to make it easier to handle. The square root of x is the same as \( x^{1/2} \). So the next step is to apply the property \( \log _{b}(a^n) = n \cdot \log _{b}(a) \). Thus, the equation simplifies to: \[2 - \frac{1}{2} \log _{8}(x+1)\]

Key Concepts

Properties of Logarithms

When working with logarithmic expressions, understanding the properties of logarithms is crucial. These properties help us simplify and manipulate expressions without the need for a calculator. Here are the key properties you need to remember:

• Product Rule: The logarithm of a product is the sum of the logarithms of its factors. That is, \( \log_b(mn) = \log_b(m) + \log_b(n) \).
• Quotient Rule: This states that the logarithm of a quotient is the difference of the logarithms. It can be written as \( \log_b(\frac{m}{n}) = \log_b(m) - \log_b(n) \).
• Power Rule: This rule indicates that the logarithm of a power is the exponent times the logarithm of the base, expressed as \( \log_b(m^n) = n \cdot \log_b(m) \).
• Change of Base Rule: A handy tool for switching the base of the logarithm to another base. It can be written as \( \log_b(m) = \frac{\log_k(m)}{\log_k(b)} \).

Learning and applying these properties will not only help you in expanding and simplifying complex logarithmic expressions but also enhance your computational efficiency.

Quotient Rule

The quotient rule of logarithms is one of the fundamental tools for simplifying logarithmic expressions. According to this rule, the logarithm of a quotient is equal to the difference between the logarithm of the numerator and the logarithm of the denominator. This is expressed mathematically as:\[\log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n)\]Let's look at an example to better understand this. Consider the logarithm expression \( \log_8\left(\frac{64}{\sqrt{x+1}}\right) \).

• The numerator is 64 and the denominator is \(\sqrt{x+1}\).
• According to the quotient rule, we can expand this expression into two separate logarithms: \( \log_8(64) - \log_8(\sqrt{x+1}) \).

This transformation is necessary to simplify further or to evaluate each part individually, helping us understand complex logarithmic expressions step by step.

Power Rule

The power rule of logarithms allows us to simplify expressions where the argument of the logarithm itself is an expression raised to a power. The power rule states that the logarithm of a number raised to a power equals the power times the logarithm of the base:\[\log_b(a^n) = n \cdot \log_b(a)\]To apply this, let's revisit the previous expression where we dealt with the square root term: \( \sqrt{x+1} \). A square root can be rewritten as an exponent, specifically \( (x+1)^{1/2} \).

• Using the power rule, \( \log_8(\sqrt{x+1}) \) can be rewritten as \( \frac{1}{2} \cdot \log_8(x+1) \).
• This makes the expression \( 2 - \frac{1}{2} \log_8(x+1) \) far easier to manipulate.

Using the power rule helps simplify logarithmic equations involving roots or other powers, allowing for neat and clearer calculations.

Base Change

Changing the base of a logarithm is useful when the calculation becomes cumbersome or when you want to express a logarithm with a different base. The change of base formula provides this flexibility. It is expressed as:\[\log_b(m) = \frac{\log_k(m)}{\log_k(b)}\]Where:

• \(b\) is the original base of the logarithm.
• \(k\) is the new base you wish to convert to, often base 10 or base \(e\) for practical reasons.

Suppose you have \( \log_8(64) \), but you need to calculate it using a base of 10. You could express this as:\[\log_{10}(64) / \log_{10}(8)\]Base change is especially handy when calculators and software only support certain bases like 10 or \(e\). It provides a straightforward method to transition between bases, ensuring accurate calculations in any context.