Problem 82 - Calculus for Biology and Medicine

Question

Simplify the following expressions: (a) \(\log _{4} 16^{x}\) (b) \(\log _{2} 16^{x}\) (c) \(\log _{3} 27^{x}\) (d) \(\log _{1 / 2} 4^{x}\) (e) \(\log _{1 / 2} 8^{-x}\) (f) \(\log _{3} 9^{-x}\)

Step-by-Step Solution

Answer

(a) 2x, (b) 4x, (c) 3x, (d) -2x, (e) -3x, (f) -2x.

Step 1: Simplify Expression (a)

We need to simplify \( \log _{4} 16^{x} \). Start by expressing 16 as \( 4^2 \), then rewrite the expression using the property of logarithms \( \log_b(a^n) = n \cdot \log_b(a) \). So, we get: \[ \log_4((4^2)^x) = \log_4(4^{2x}) = 2x \cdot \log_4(4). \] Since \( \log_4(4) = 1 \), the expression simplifies to \( 2x \).

Step 2: Simplify Expression (b)

In \( \log _{2} 16^{x} \), start by rewriting 16 as \(2^4\). Then apply the properties of logarithms: \[ \log_2((2^4)^x) = \log_2(2^{4x}) = 4x \cdot \log_2(2). \] Since \( \log_2(2) = 1 \), the expression simplifies to \( 4x \).

Step 3: Simplify Expression (c)

For \( \log _{3} 27^{x} \), rewrite 27 as \( 3^3 \): \[ \log_3((3^3)^x) = \log_3(3^{3x}) = 3x \cdot \log_3(3). \] Since \( \log_3(3) = 1 \), the expression simplifies to \( 3x \).

Step 4: Simplify Expression (d)

To simplify \( \log _{1 / 2} 4^{x} \), express 4 as \((\frac{1}{2})^{-2}\): \[ \log_{1/2}((\frac{1}{2})^{-2x}) = -2x \cdot \log_{1/2}(\frac{1}{2}). \] Here, \( \log_{1/2}(\frac{1}{2}) = 1 \), so the expression simplifies to \( -2x \).

Step 5: Simplify Expression (e)

For \( \log _{1 / 2} 8^{-x} \), write 8 as \((\frac{1}{2})^{-3}\) and simplify: \[ \log_{1/2}((\frac{1}{2})^{-3x}) = -3x \cdot \log_{1/2}(\frac{1}{2}). \] Since \( \log_{1/2}(\frac{1}{2}) = 1 \), the expression simplifies to \( -3x \).

Step 6: Simplify Expression (f)

Simplify \( \log _{3} 9^{-x} \) by expressing 9 as \(3^2\): \[ \log_3((3^2)^{-x}) = \log_3(3^{-2x}) = -2x \cdot \log_3(3). \] Since \( \log_3(3) = 1 \), the expression simplifies to \( -2x \).

Key Concepts

Logarithm Properties

Logarithms are a fundamental mathematical concept, used to relate exponents in a variety of ways. One of their key properties includes how they handle powers: \( \log_b(a^n) = n \cdot \log_b(a) \). This means that if you have a power inside a logarithm, you can "bring down" the exponent in front of the log.

For example, in the expression \( \log_4((4^2)^x) \), you can reframe this as \( \log_4(4^{2x}) \). Using the aforementioned property, the expression can be simplified to \( 2x \cdot \log_4(4) \).

Another important property involves the base of the logarithm: \( \log_b(b) = 1 \) for any positive number \( b \). This is because the logarithm essentially counts how many times we multiply the base \( b \) to get itself. As a result, expressions like \( \log_4(4) \) simplify to 1.

In summary, by understanding these properties, you're equipped to efficiently simplify logarithmic expressions involving powers, leading to much cleaner, simpler results.

Simplification Techniques

Simplifying logarithmic expressions often involves expressing numbers in terms of the base of the logarithm. This can make complex expressions much easier to handle.

Consider the expression \( \log_{2}(16^x) \). The number 16 can be expressed as \( 2^4 \). Thus, \( \log_2((2^4)^x) \) transforms to \( \log_2(2^{4x}) \).

• Apply the power property: \( 4x \cdot \log_2(2) \).
• Recognize that \( \log_2(2) = 1 \).

The expression simplifies to \( 4x \).

This technique not only applies to positive bases but also to fractions. For example, the expression \( \log_{1/2}(4^x) \) becomes easier once you recognize \( 4 \) as \( (1/2)^{-2} \). Simplifying using properties as before, you achieve a much simpler outcome.

The core of simplification lies in transforming numbers into a common base with the logarithm, employing properties to adjust the expression, and then reducing redundancies.

Base Conversion in Logarithms

Changing the base of a logarithm can sometimes simplify calculations or make them more applicable to a specific problem. A common change involves converting the base using the formula: \[ \log_a(b) = \frac{\log_c(b)}{\log_c(a)} \] Here, \( c \) is the new base you want to switch to. This conversion formula explains how large or small a difference exists between bases.

While converting bases wasn't needed for the original exercise, it's a helpful technique when dealing with logs that do not share a common base with their numbers. For example, if you had a situation involving a tricky base, such as \( \log_3(81) \), it may sometimes help to switch to a common base like \( 2 \) or \( 10 \).

Remember, understanding how to change bases gives you a powerful tool in algebra, letting you manipulate expressions to fit your needs better. This flexibility proves invaluable in advanced calculations and ensures you aren't stuck when faced with unfamiliar bases.