Problem 25 - Algebra and Trigonometry

Question

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 expanded and simplified form of \( \log _{6}\left(\frac{36}{\sqrt{x+1}}\right) \) is \( 2 - \frac{1}{2}\log_{6}(x+1) \)

Step 1: Apply the Logarithmic Quotient Property

The logarithmic quotient property states that the log of a quotient is the difference of the logs. So, from this property, we can write \( \log _{6}\left(\frac{36}{\sqrt{x+1}}\right) \) as \( \log_{6}(36) - \log_{6}(\sqrt{x+1}) \).

Step 2: Evaluate \( \log_{6}(36) \) and Apply the Logarithmic Square Root Property

The \( \log_{6}(36) \) can be simplified as 2 because 6 raised to the 2nd power equals 36. For \( \log_{6}(\sqrt{x+1}) \), apply the square root property of logarithms, which states that the log of a square root is half the log. This allows us to transform \( \log_{6}(\sqrt{x+1}) \) into \( \frac{1}{2}\log_{6}(x+1) \). So, the expression simplifies to \( 2 - \frac{1}{2}\log_{6}(x+1) \).

Step 3: Final Form

By executing the properties precisely, the original logarithmic expression expands and simplifies to \( 2 - \frac{1}{2}\log_{6}(x+1) \)

Key Concepts

Logarithmic Quotient Property

Understanding the logarithmic quotient property is like having a mathematical Swiss Army knife for dealing with divisions within a logarithm. This property simply states that the log of a quotient is equal to the difference of the logs of the numerator and the denominator. In mathematical terms, for any positive numbers 'a', 'b', and base 'c' that are not equal to 1, the property is expressed as:
\[ \log_{c}\left(\frac{a}{b}\right) = \log_{c}(a) - \log_{c}(b) \]
Applying this property makes it easier to simplify complex logarithmic expressions and can also be invaluable when solving logarithmic equations. When faced with a problem like \(\log_{6}\left(\frac{36}{\sqrt{x+1}}\right)\), recognizing that you can split the log of the quotient into the difference of two logs streamlines the process and sets the stage for further simplification.

Logarithmic Square Root Property

Similar to how the quotient property helps us with division inside a logarithm, the logarithmic square root property assists us when dealing with roots, particularly square roots. This property says that the logarithm of a square root is half of the logarithm of the number under the root. To put it into a formula, for any positive number 'a' and base 'c':
\[ \log_{c}(\sqrt{a}) = \frac{1}{2}\log_{c}(a) \]
This understanding is crucial for simplifying problems like \(\log_{6}(\sqrt{x+1})\) where this property converts our expression into something more straightforward. By acknowledging that the square root can be represented as an exponent of 1/2, we simplify the original problem, turning \(\log_{6}(\sqrt{x+1})\) into \(\frac{1}{2}\log_{6}(x+1)\).

Evaluating Logarithms

Evaluating logarithms can sometimes seem as daunting as deciphering an ancient code. However, once you know the properties and can identify the base and the argument, the task becomes much more manageable. To evaluate a logarithm, such as \(\log_{6}(36)\), you're essentially asking 'to what exponent must the base (6) be raised in order to produce the argument (36)?'. In this particular case, because \(6^2 = 36\), the logarithm evaluates to 2.
Understanding the basic properties of logarithms enables you to break down more complex expressions into simpler pieces that you can evaluate. It's important to remember that the result of a logarithm will always be the exponent, which often requires a bit of mental or written calculation when the logarithm doesn't simplify to a common base. Nevertheless, with practice and the right approach, evaluating logarithms can become a standard procedure in one's mathematical toolkit.