Problem 105 - College Algebra

Question

If \(f(x)=\log _{\phi} x,\) show that \(\frac{f(x+h)-f(x)}{h}=\log _{b}\left(1+\frac{h}{x}\right)^{1 / h}, h \neq 0\)

Step-by-Step Solution

Answer

The result of the limit of the difference quotient \( \frac{f(x+h)-f(x)}{h} \) simplifies to \( \log _{\phi} e = 1 \)

Step 1: Rearrange terms

Rewrite the difference quotient as a single logarithm by: \( f(x+h)=\log _{\phi}(x+h) = \log _{\phi} x+\log _{\phi} (1+\frac{h}{x})\). Thus the difference quotient can be written as \( \frac{f(x+h)-f(x)}{h}=\frac{\log _{\phi} (1+\frac{h}{x}) } {h}\).

Step 2: Use change of base formula

Applying the change of base formula with base \(\phi\) enables us to write this in term of natural logarithm. The formula \( \log_b a = \frac{\ln a}{\ln b} \) yields \( \frac{\log _{\phi} (1+\frac{h}{x})} {h}=\frac{\ln (1+\frac{h}{x})} {h \ln \phi}\).

Step 3: Apply the limit

As \(h\) approaches \(0\), the expression \( \ln(1+\frac{h}{x})^ \frac{1}{h} \) will result in \(\ln e\), which is \(1\) because \( e = \lim_{h \to 0}\left(1+\frac{h}{x}\right)^\frac{x}{h} \). Thus, \( \frac{\ln (1+\frac{h}{x})} {h \ln \phi} \) simplifies to \(\frac{1}{\ln \phi}\) or \( \log _{\phi} e = 1 \).

Key Concepts

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions. They are crucial in various scientific fields, including physics, chemistry, and engineering, to solve problems involving exponential growth or decay. The general form of a logarithm is written as \( \log_b x \), where \(b\) is the base and \(x\) is the argument of the logarithm. For instance, in the exercise, \(f(x) = \log_{\phi} x\) represents a logarithmic function with base \(\phi\).

It is important to understand the behavior of these functions, especially how they translate addition into multiplication and vice versa. For example, \( \log_b(xy) = \log_b(x) + \log_b(y) \). In the given exercise, we use this property to express the difference quotient as a single logarithm, which is essential for finding derivatives and understanding rates of change in calculus.

Change of Base Formula

The change of base formula is a pivotal concept in working with logarithms, especially when the base of the logarithm in question isn't particularly convenient for the problem we're solving. This formula allows us to convert a logarithm with any base to a logarithm with a different base, which is often chosen to simplify calculations.

The formula is given by: \( \log_{b} a = \frac{\ln a}{\ln b} \), where \( \ln \) denotes the natural logarithm (logarithm with base \(e\)). So in our exercise, we changed the base from \( \phi \) to \( e \) using this formula, which makes it easier to handle the limit as \(h\) approaches 0. With this tool, we transitioned from a potentially unfamiliar base \(\phi\) to a more common base that aligns with the exponential function \( e^x \), a cornerstone in the study of calculus.

Limits in Calculus

Limits are fundamental in calculus and provide a way to deal with values that approach a certain number but never quite reach it. They are used to define the most basic building blocks of calculus: derivatives and integrals. A limit evaluates what happens to a function as its input \(x\) approaches a particular value, and this understanding is essential to determine continuity, rates of change, and the behavior of functions on a graph.

In our exercise, we apply the limit as \(h \) approaches zero to the function \( \ln(1+\frac{h}{x})^ \frac{1}{h} \), which simplifies to \( \log _{\phi} e \) or \( 1 \) after some manipulation. This transformation demonstrates the concept that as \(h\) becomes infinitesimally small, the difference quotient reflects the instantaneous rate of change of \(\log_{\phi} x\) at a point, which is the key idea behind the derivative in calculus.