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. $$ \log _{x}\left(\frac{64}{\sqrt{x+1}}\right) $$

Step-by-Step Solution

Answer

The result of expanding the logarithmic expression \(\log _{x}\left(\frac{64}{\sqrt{x+1}}\right)\) as much as possible is \(6 \cdot \log _{x}(2) - \frac{1}{2}\)

Step 1: Apply the Quotient Rule of Logarithms

The quotient rule of logarithms states that \( log_b(\frac{M}{N}) = log_b(M)-log_b(N) \), so we can write the given expression as follows: \(\log _{x}(64) - \log _{x}(\sqrt{x+1}) \)

Step 2: Apply the Power Rule of Logarithms

Since \(64 = 2^{6}\), the power rule of logarithms that states \(log_b(M^n) = n \cdot log_b(M) \) is applied to \(\log _{x}(64) \) to simplify our expression: \(6 \cdot \log _{x}(2) - \log _{x}(\sqrt{x+1}) \)

Step 3: Apply the Square Root Rule of Logarithms to Further Simplify

As the rule \( log_b(\sqrt[n]{M}) = \frac{1}{n} \cdot log_b(M) \) suggests, the square root inside the log expression in the second part of the equation could be brought out in front of the log as a factor to obtain: \(6 \cdot \log _{x}(2) - \frac{1}{2} \cdot \log _{x}(x+1) \)

Step 4: Final Simplification

Let's go one step further to simplify this expression. Given that \(\log _{x}(x) = 1\), our equation simplifies to: \(6 \cdot \log _{x}(2) - \frac{1}{2} \)

Key Concepts

Quotient Rule of Logarithms

The quotient rule of logarithms is a handy tool when dealing with divisions inside a log. It helps transform a fraction within a logarithm into a subtraction outside of it. This is especially useful as subtraction is often easier to handle than division.

• Take the fraction inside the log: \( \log_x\left(\frac{M}{N}\right) \)
• Turn it into subtraction: \( \log_x(M) - \log_x(N) \)

In the original exercise, this rule was applied to \( \log _{x}\left(\frac{64}{\sqrt{x+1}}\right) \), so it became two separate terms: \( \log _{x}(64) - \log _{x}(\sqrt{x+1}) \). This step lays the groundwork for further simplification, making complex expressions a bit simpler to work with.

Power Rule of Logarithms

The power rule of logarithms is invaluable when a logarithm involves an exponent. This rule allows us to take an exponent and move it in front of the logarithm as a multiplier.

• If you have \( log_b(M^n) \), then it becomes \( n \cdot log_b(M) \).
• This makes complicated powers much simpler by reducing them to basic multiplication.

In the step-by-step solution, transforming \( 64 \, (64 = 2^6) \) was an ideal situation to use this rule. The log expression \( \log _{x}(64) \) was converted using the power rule into \( 6 \cdot \log _{x}(2) \). By doing this, we efficiently simplify expressions involving powers, making calculations manageable.

Square Root in Logarithms

Whenever you encounter a square root within a logarithm, there's a straightforward way to simplify it. This involves recognizing the square root as a power and using the properties like in power rules.

• Recall that \( \sqrt{M} \) is the same as \( M^{1/2} \).
• Use the rule: \( log_b(\sqrt[n]{M}) = \frac{1}{n} \cdot log_b(M) \).

In this exercise, \( \log _{x}(\sqrt{x+1}) \) was simplified using the square root rule. It became \( \frac{1}{2} \cdot \log _{x}(x+1) \). The conversion simplifies the expression, allowing us to handle roots just like we do with multiplication and division. Such simplifications make understanding and working with logarithms more intuitive.