Problem 29 - Precalculus

Question

Use the properties of logarithms to write the expression as a single logarithm. $$ \log _{2}(x)+\log _{\frac{1}{2}}(x-1) $$

Step-by-Step Solution

Answer

\(\log_{2}\left(\frac{x}{x-1}\right)\)

Step 1: Apply the Change of Base Formula

To combine the logarithms with different bases, we first apply the change of base formula. The change of base formula is: \[\log_{b}(a) = \frac{\log_{c}(a)}{\log_{c}(b)}\] We can use any common base, but let's use base 2 for simplicity. - First, convert \(\log_{\frac{1}{2}}(x-1)\) to base 2: \[ \log_{\frac{1}{2}}(x-1) = \frac{\log_{2}(x-1)}{\log_{2}(\frac{1}{2})} \] - Since \(\log_{2}(\frac{1}{2}) = \log_{2}(2^{-1}) = -1\), this simplifies to: \[ \log_{\frac{1}{2}}(x-1) = -\log_{2}(x-1) \]

Step 2: Combine Logarithms

Now that both logarithms are in base 2, we can combine them using the properties of logarithms. The property we use is:\[\log_{b}(a) + \log_{b}(c) = \log_{b}(a \times c)\]Rewriting the expression, we get:\[\log_{2}(x) + \log_{\frac{1}{2}}(x-1) = \log_{2}(x) + (-\log_{2}(x-1)) = \log_{2}(x) - \log_{2}(x-1)\]Using the property of logarithms, we can express this as a single logarithm:\[\log_{2}\left(\frac{x}{x-1}\right)\]

Step 3: Write the Final Expression

The expression \(\log_{2}(x) + \log_{\frac{1}{2}}(x-1)\) is equivalent to a single logarithm:\[\log_{2}\left(\frac{x}{x-1}\right)\] This is the simplest form of the expression using the properties of logarithms.

Key Concepts

Change of Base Formula

The change of base formula is a very useful tool in dealing with logarithms. Sometimes, you might encounter logarithms that are written in bases other than the one you're working with or familiar with. To convert a logarithm with an unfamiliar base to one that is easier to work with, use this formula:

• \( \log_{b}(a) = \frac{\log_{c}(a)}{\log_{c}(b)} \)

This lets you rewrite a logarithm in a new base, \(c\). The choice of base \(c\) is usually up to you; however, common choices are 10 (for common logarithms) or \(e\) (for natural logarithms). For simplicity, other common bases like 2 can also be used very often, especially in exercises working with binary concepts.

In our example, we needed to convert \(\log_{\frac{1}{2}}(x-1)\) into a base that aligns with another term, which was base 2. After applying the change of base formula, it was effectively transformed to \(-\log_{2}(x-1)\), making both terms we are working with in base 2, ready to be combined.

Combining Logarithms

Once you're working with logarithms of the same base, it's time to simplify them using the properties of logarithms. These properties help in combining or breaking down logarithms. Here’s the essential property you need:

• \( \log_{b}(a) + \log_{b}(c) = \log_{b}(a \times c) \)

This property shows how logarithms that are added can be combined into a single logarithm by multiplying their inside terms (the arguments). There’s a similar property for subtraction:

• \( \log_{b}(a) - \log_{b}(c) = \log_{b}\left(\frac{a}{c}\right) \)

This helps express a subtraction of logarithms as a division of their arguments.

Our expression from the exercise becomes \(\log_{2}(x) - \log_{2}(x-1)\). Using the subtraction property, it simplifies down to \(\log_{2}\left(\frac{x}{x-1}\right)\), making it a single logarithm expression.

Logarithmic Expressions

Logarithmic expressions can sometimes seem complicated, but with the right tools, you can simplify them and make them much easier to manage. These expressions often involve the use of various algebraic properties and rules specific to logarithms.

Remember some key points:

• Use the change of base formula to convert logarithms into a common base when necessary.
• Combine logarithms using addition and subtraction rules accordingly to consolidate them into simpler forms.
• When simplified, these expressions can help solve equations or simplify complex mathematical models in real-world scenarios.

The core of working with logarithmic expressions is recognizing these rules and applying them effectively, which not only simplifies problems but can also make seemingly complex challenges more approachable.