Problem 25 - 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. $$ \log _{6}\left(\frac{36}{\sqrt{x+1}}\right) $$

Step-by-Step Solution

Answer

The logarithmic expression simplifies to \(2 - \frac{1}{2} \cdot \frac{\log (x+1)}{\log 6}\)

Step 1: Apply Quotient Rule

The quotient rule of logarithms states that \( \log_b \left(\frac{m}{n}\right) \) equals \( \log_b (m) - \log_b (n) \). Therefore, we can express \(\log _{6}\left(\frac{36}{\sqrt{x+1}}\right)\) as \(\log _{6}(36) - \log _{6}(\sqrt{x+1})\)

Step 2: Apply Change of Base Formula

The change of base formula states that \( \log_b a = \frac{\log a}{\log b} \). We know that \( \log _{6}36 = 2 \) because 6 squared equals 36. Therefore, this part of the expression is 2 and doesn't need further simplification.

Step 3: Simplify Logarithm with Root

For \( \log _{6}(\sqrt{x+1})\), we can apply the rule that states \( \log \sqrt[n]{b} = \frac{1}{n} \log b \) . Therefore, \( \log _{6}(\sqrt{x+1}) = \frac{1}{2} \log _{6}(x+1)\). We can apply the change of base formula to get \( \log _{6}(\sqrt{x+1}) = \frac{1}{2} \cdot \frac{\log (x+1)}{\log 6} \).

Step 4: Putting It All Together

Now, substitute the results from Steps 2 and 3 back into the expression we got in Step 1: \(2 - \frac{1}{2} \cdot \frac{\log (x+1)}{\log 6}\). This is the final expanded form of the initial logarithmic expression.