Problem 80 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

(a) \(\log_{3} 4\), (b) \(\log x + \log(x+1)\), (c) \(x - 2\).

Step 1: Simplify Expression (a)

The expression is \( -\log_{3}\frac{1}{4} \). According to the properties of logarithms, \(\log_{b}\frac{1}{a} = -\log_{b}a\). So, we can write: \[ -\log_{3} \frac{1}{4} = \log_{3} 4 \] Thus, the simplified form of expression (a) is \( \log_{3} 4 \).

Step 2: Simplify Expression (b)

The given expression is \( \log \left(\frac{x^{3}-x}{x-1}\right) \). First, simplify the fraction: \( x^3 - x = x(x^2 - 1) = x(x-1)(x+1)\) and \( x - 1 \).Therefore, \( \frac{x^3-x}{x-1} = x(x+1) \).Applying the property of logarithms, \( \log ab = \log a + \log b \), we get: \[ \log \left(x(x+1)\right) = \log x + \log(x+1) \] So, expression (b) simplifies to \( \log x + \log(x+1) \).

Step 3: Simplify Expression (c)

The expression is \( \ln \left(e^{x-2}\right) \). Using the property \( \ln(e^a) = a \), the expression simplifies directly to: \[ x - 2 \] Thus, the simplified form for expression (c) is \( x - 2 \).

Key Concepts

Properties of Logarithms

Logarithms have special properties that allow us to transform and simplify expressions, making them more manageable. One key property is the 'quotient rule', which states that the logarithm of a quotient is equal to the difference of the logarithms:

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

Another useful property is the 'product rule'. This property suggests that the logarithm of a product is the sum of the logarithms of the factors:

• \( \log_b (ac) = \log_b a + \log_b c \)

These rules are incredibly helpful when dealing with products or quotients inside the logarithm. In part (a) of the original exercise, the transformation from a reciprocal to a positive logarithm was achieved using the inverse of the quotient rule. Recognizing and applying these properties correctly is crucial for simplifying logarithmic expressions.

Simplifying Algebraic Expressions

Simplifying algebraic expressions is the process of making expressions less complex, which is done by factoring and canceling out common terms. For the expression given in part (b), \( \log \left(\frac{x^{3}-x}{x-1}\right) \), simplifying begins with factoring the terms in the numerator. The expression \( x^3 - x \) can be factored as \( x(x^2 - 1) \), and further as \( x(x-1)(x+1) \). This reveals common factors with the denominator. By canceling the \( x - 1 \) terms, the expression turns into \( x(x+1) \).
Then, applying the properties of logarithms, specifically the 'product rule', aids in further simplification to \( \log x + \log(x+1) \). Step-by-step simplification minimizes the complexity and makes the expression easier to work with in further calculations or evaluations. Recognizing and detaching factors are essential skills for managing algebraic expressions.

Natural Logarithms

Natural logarithms, often denoted as \( \ln \), are logarithms with a base of \( e \), which is an important mathematical constant approximately equal to 2.71828. A key property of natural logarithms involves their relationship with the exponential function. Specifically, the property \( \ln(e^a) = a \) allows for straightforward simplification of expressions involving \( e \).
In the context of expression (c), \( \ln \left(e^{x-2}\right) \), the property directly simplifies the expression to \( x - 2 \). This property is especially powerful because it effectively "cancels out" the exponential, leaving just the exponent.

• This leads to quick simplifications when working with exponential growth, decay, or continuous compounding in calculus or finance.

Recognizing \( e \) and \( \ln \) as inverse operations is essential when handling natural logarithms, making computations straightforward and efficient.