Problem 79 - Calculus for Biology and Medicine

Question

Simplify the following expressions: (a) \(-\ln \frac{1}{3}\) (b) \(\log _{4}\left(x^{2}-4\right)\) (c) \(\log _{2} 4^{3 x-1}\)

Step-by-Step Solution

Answer

(a) \(\ln(3)\), (b) \(\log_{4}(x-2) + \log_{4}(x+2)\), (c) \(6x - 2\).

Step 1: Understanding Negative Logarithm

To simplify \(-\ln \frac{1}{3}\), you should understand the property of logarithms: \(\ln\left(\frac{1}{a}\right) = -\ln(a)\). Use this principle to invert the fraction inside the logarithm.

Step 2: Applying Logarithm Properties

Using the property \(\ln\left(\frac{1}{3}\right) = -\ln(3)\), simplify the expression. Therefore, \(-\ln \frac{1}{3} = \ln(3)\).

Step 3: Factoring Quadratic Expression

For expression \(\log _{4}(x^2 - 4)\), notice that the quadratic expression \(x^2 - 4\) can be factored into \( (x-2)(x+2)\), which is a difference of squares.

Step 4: Logarithmic Split Using Product Rule

With the factorization, the product inside the logarithm can be split using \(\log_b(mn) = \log_b(m) + \log_b(n)\). Thus, \(\log _{4}(x^2 - 4) = \log_{4}(x-2) + \log_{4}(x+2)\).

Step 5: Simplifying Power Expression

In \(\log _{2} 4^{3x-1}\), recognize that \(4\) is a power of \(2\), \(4 = 2^2\). Substitute to get \(\log_2 ((2^2)^{3x-1})\).

Step 6: Using Power Rule of Logarithms

Apply the power rule: \(\log_b(m^n) = n\cdot\log_b(m)\). This gives \(\log_2 ((2^2)^{3x-1}) = (3x-1)\cdot\log_2(2^2) = 2(3x-1)\), because \(\log_2(2) = 1\).

Step 7: Final Simplification

Now simplify \(2(3x-1)\) to get \(6x - 2\). Thus, \(\log_2 4^{3x-1} = 6x - 2\).

Key Concepts

Understanding Logarithms

Logarithms are a fundamental concept in mathematics, particularly in calculus. They are the inverse operations of exponentiation, which means they help us solve the question: "To what power must a given base be raised, to produce a certain number?"
Consider the equation \ y = \log_b(x) \, which means that base \( b \) raised to the power of \( y \) produces \( x \). Logarithms are particularly useful when dealing with exponential growth or decay, and they have properties that make them easier to manipulate.

Key properties of logarithms include:

• Product Rule: \( \log_b(mn) = \log_b(m) + \log_b(n) \)
• Quotient Rule: \( \log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n) \)
• Power Rule: \( \log_b(m^n) = n \cdot \log_b(m) \)
• Change of Base Formula: \( \log_b(m) = \frac{\log_k(m)}{\log_k(b)} \) for any positive base \( k \)

Understanding these rules is crucial in simplifying logarithmic expressions, as seen in the original exercise where negative properties and inversion of fractions for logarithms were used to transform \(-\ln \left(\frac{1}{3}\right)\) into \(\ln(3)\).

Simplifying Expressions

Expression simplification involves rewriting expressions in a simpler or more manageable form, and is a skill often required in calculus.
It involves using various algebraic techniques such as factoring, expanding, and canceling.

In mathematical contexts like above where \(x^2 - 4\) appears within a logarithm, recognizing common algebraic identities can help streamline the simplification process.

A widely used technique is the "difference of squares," where an expression of the form \(a^2 - b^2\) can be rewritten as \((a-b)(a+b)\). This approach was applied to rewrite \(x^2 - 4\) as \((x-2)(x+2)\), allowing the log expression mation \(\log_4(x^2 - 4)\) to be split using the product rule of logarithms.

By mastering these foundational techniques, students can tackle complex expressions with more confidence, ensuring each step of the calculation brings them closer to a simplified answer.

The Power Rule of Logarithms

The power rule of logarithms is a handy tool that simplifies expressions involving powers of a base. It is expressed as \( \log_b(m^n) = n \cdot \log_b(m) \). This rule essentially allows you to "bring down" the exponent as a coefficient, making it easier to handle exponential terms.

This rule is invaluable when simplifying logarithmic expressions that involve raised bases. For instance, in the given problem \(\log_2 (4^{3x-1})\), recognizing that \(4\) is equivalent to \(2^2\) allowed us to quickly apply the power rule.
By substituting \(4\) with \((2^2)\), we transformed the problem into \(\log_2((2^2)^{3x-1})\).

Applying the power rule, we could factor the exponent outside, leading to \((3x-1) \cdot \log_2(2^2)\). Since \(\log_2(2) = 1\), this simplifies down to \(2(3x-1)\) and thus \(6x - 2\).
This process underscores the power of the logarithmic rules in breaking down intricate functions into manageable components, ultimately leading to simplified forms.