Problem 45 - Algebra A Combined Function

Question

Write each expression as a sum or difference of logarithms. Assume that variables represent positive numbers. $$ \log _{b} \sqrt{7 x} $$

Step-by-Step Solution

Answer

The expression is \( \frac{1}{2} \log_{b} 7 + \frac{1}{2} \log_{b} x \).

Step 1: Rewrite the Radical as an Exponent

Recognize that the square root of a number can be expressed as an exponent of 1/2. Therefore, rewrite the expression inside the logarithm:\[ \log_{b} \sqrt{7x} = \log_{b} (7x)^{1/2} \]

Step 2: Apply the Power Rule for Logarithms

Use the logarithm power rule, which states \( \log_{b} (M^n) = n \cdot \log_{b} M \). Apply this rule to the expression:\[ \log_{b} (7x)^{1/2} = \frac{1}{2} \cdot \log_{b} (7x) \]

Step 3: Apply the Product Rule for Logarithms

Use the product rule for logarithms, which states \( \log_{b} (MN) = \log_{b} M + \log_{b} N \). Apply this to \( \log_{b} (7x) \):\[ \log_{b} (7x) = \log_{b} 7 + \log_{b} x \]

Step 4: Combine the Expressions

Combine the results from Steps 2 and 3:\[ \frac{1}{2} \cdot \log_{b} (7x) = \frac{1}{2} \cdot (\log_{b} 7 + \log_{b} x) \]Now distribute the \( \frac{1}{2} \):\[ \frac{1}{2} \cdot \log_{b} 7 + \frac{1}{2} \cdot \log_{b} x \]

Key Concepts

Logarithm Properties

Logarithm properties make complicated expressions much easier to work with by simplifying or expanding them. This is especially helpful when solving equations or comparing values. Here are some key properties to remember:

• Logarithm of a product: The log of a product is the sum of the logs: \(\log_{b}(MN) = \log_{b} M + \log_{b} N\).
• Logarithm of a quotient: The log of a quotient is the difference of the logs: \(\log_{b}\left(\frac{M}{N}\right) = \log_{b} M - \log_{b} N\).
• Logarithm of a power: The log of a power brings the exponent down as a multiplier: \(\log_{b}(M^n) = n \cdot \log_{b} M\).
• Log of 1: Any base raised to the power of 0 equals 1, so \(\log_{b} 1 = 0\).
• Log equals base: If the number equals the base, then the log is 1, such as \(\log_{b} b = 1\).

These properties are the toolkit for manipulating and understanding logarithmic expressions.

Product Rule for Logarithms

The product rule is essential for breaking down logarithms involving multiplication into simpler, more manageable parts. This rule states: \[\log_{b}(MN) = \log_{b} M + \log_{b} N\]Imagine you have a log of a product like \(\log_{b}(7x)\). According to the product rule, you can write this as \(\log_{b} 7 + \log_{b} x\). This is useful because it transforms a single logarithm into a sum, making it easier to handle individual components.
Each separate log \(\log_{b} 7\) and \(\log_{b} x\) can often be solved or used in calculations individually, providing a clearer insight into the problem.

Power Rule for Logarithms

The power rule is a powerful tool when dealing with logarithms raised to an exponent. It allows you to bring the exponent to the front, making the expression easier to work with. This rule states:\[\log_{b}(M^n) = n \cdot \log_{b} M\]For example, in the expression \(\log_{b}((7x)^{1/2})\), the exponent \(1/2\) is moved out in front of the log as a multiplier. This gives us \(\frac{1}{2} \cdot \log_{b}(7x)\).
This manipulation simplifies handling powers by turning a complex exponential expression into a much simpler arithmetic operation. Thus, when you encounter an exponent in a log, the power rule should be your go-to strategy.

Radicals and Exponents

Understanding the connection between radicals and exponents is pivotal when working with logarithms. Radicals can be expressed as fractional exponents, which can simplify expressions greatly. For instance:

• The square root of a number \(\sqrt{x}\) can be written as \(x^{1/2}\).
• The cube root of a number \(\sqrt[3]{x}\) becomes \(x^{1/3}\), and so on.

So when you see \(\log_{b}(\sqrt{7x})\), you transform it into \(\log_{b}((7x)^{1/2})\). This step allows you to use the power rule easily.
Remember, recognizing these energy-efficient transformations between radicals and exponents is vital for simplifying logarithmic expressions and ultimately mastering their calculations.