Problem 25 - College Algebra Essentials

Question

Use properties of logarithms to expand logarithmic expression as much as possible. Where possible, evaluate logarithmic expressions without using a calculator. \(\log _{6}\left(\frac{36}{\sqrt{x+1}}\right)\)

Step-by-Step Solution

Answer

The expanded form of the logarithmic expression \(\log _{6}\left(\frac{36}{\sqrt{x+1}}\right)\) is \(2 - \frac{1}{2} \log _{6} (x + 1)\).

Step 1: Identify the properties of logarithms

We have three core properties of logarithms that will be useful: 1) \(\log_b(MN) = \log_b(M) + \log_b(N)\); 2) \(\log_b(M/N) = \log_b(M) - \log_b(N)\); and 3) \(\log_b(M^n) = n \cdot \log_b(M)\). These properties will help break down the complex logarithmic expression.

Step 2: Apply the division property of logarithms

According to the division property of logarithms, we can split the log into subtraction of two logs: \(\log _{6} 36 - \log _{6} \sqrt{x + 1}\). This is derived from the formula \(\log_b(M/N) = \log_b(M) - \log_b(N)\).

Step 3: Simplify the first logarithm

The first part, \(\log _{6} 36\), can be solved without a calculator as 6 squared gives 36. Hence, \(\log _{6} 36 = 2\)

Step 4: Apply the power property to the second logarithm

For the second part, \(\log _{6} \sqrt{x + 1}\), we can apply the power property of logarithms. The \(\sqrt{x + 1}\) term is equivalent to \({(x + 1)}^{1/2}\), so this can be rewritten as \(\frac{1}{2} \log _{6} (x + 1)\) based on the property \(\log_b(M^n) = n \cdot \log_b(M)\).

Step 5: Final expanded form

Putting all together, we end up with \(2 - \frac{1}{2} \log _{6} (x + 1)\)

Key Concepts

Properties of Logarithms

Logarithms have special properties that make complex expressions simpler to work with. When working with logarithmic expressions, you frequently come across three essential properties:

• Product Property: This allows you to split the logarithm of a product into a sum. That is, \( \log_b(MN) = \log_b(M) + \log_b(N)\).
• Quotient Property: It turns the logarithm of a division into a difference. This is expressed as \( \log_b(M/N) = \log_b(M) - \log_b(N)\).
• Power Property: It moves an exponent in the argument into a multiplier. So, \( \log_b(M^n) = n \cdot \log_b(M)\).

Each property helps in expanding and simplifying logarithmic expressions. By applying these properties appropriately, you can take a complicated expression and break it down into more manageable parts.

Expansion of Logarithmic Expressions

Expanding a logarithmic expression means expressing it as a sum, difference, and/or multiples of simple logarithms. Let's illustrate this using the example: \( \log _{6}\left(\frac{36}{\sqrt{x+1}}\right) \).

Firstly, using the Quotient Property, divide into separate logs: \( \log_{6} 36 - \log_{6} \sqrt{x + 1} \). This helps split a complex fraction into two parts, one log being subtracted from the other.

Next, evaluate each part individually wherever possible. The first term, \( \log_{6} 36 \), simplifies using integer powers because 6-squared equals 36, so the value is 2.

For the second term, apply the Power Property. The square root can be rewritten in exponential form: \( \sqrt{x+1} = (x+1)^{1/2} \). Thus, the expression becomes \( \frac{1}{2} \log_{6} (x+1) \) after using the power property.

Combine all to form the expansion:\( 2 - \frac{1}{2} \log_{6} (x+1) \). This demonstrates how to break down and reassemble a complex logarithmic term.

Evaluating Logarithms without a Calculator

Not all logarithms require a calculator for evaluation, especially when you recognize patterns or identities. For instance, consider \( \log_{6} 36 \). To evaluate without a calculator, rearrange the potential powers of 6:

Since \( 6^2 = 36 \), apply the rule that \( \log_b (b^n) = n \). Therefore, \( \log_{6} 36 = 2 \) because the base 6 raised to the power of 2 yields 36.

Being able to work out logarithms in this manner saves time and sharpens your mental math skills. Look for integer powers of the base that match the argument, or use known identities that simplify expressions directly.

These skills are handy when dealing with standardized tests or situations where calculators are not permitted. Recognizing these quick evaluations will help boost your confidence in handling logarithmic expressions effortlessly.