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Chapter 4: Problem 102

$$ y=\ln \left(\frac{e^{x}-1}{e^{x}}\right) $$

Short Answer

Expert verified
The given function \(y = \ln\left(\frac{e^x -1}{e^x}\right)\) can be simplified as follows: Divide both terms in the numerator by \(e^x\) to obtain \(1 - \frac{1}{e^x}\). Substituting this simplified expression back into the logarithm gives the final simplified function: \(y = \ln\left(1 - \frac{1}{e^x}\right)\).

Step by step solution

01

Identify the logarithmic and exponential functions

In this exercise, the function is given as \(y = \ln\left(\frac{e^x -1}{e^x}\right)\). We can see that it involves both a logarithmic function (with base e) and an exponential function (also with base e).
02

Simplify the fraction inside the logarithm

Let's simplify the fraction inside the logarithm: \[\frac{e^x -1}{e^x}.\] To do this, divide both terms in the numerator by \(e^x\): \[\frac{e^x}{e^x} - \frac{1}{e^x},\] which simplifies to: \[1 - \frac{1}{e^x}.\] Now, we have the simplified expression inside the logarithm:
03

Substitute the simplified expression back into the logarithm

Now that we've simplified the fraction, we can plug it back into the logarithm. Our function now becomes: \[y = \ln\left(1 - \frac{1}{e^x}\right).\] This is our final, simplified expression for the given function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithmic functions
Logarithmic functions are an important concept in calculus, particularly when dealing with exponential growth and decay. Essentially, a logarithm is the inverse operation of exponentiation. In simpler terms, while exponentiation involves raising a number to a power, a logarithm is used to determine what power a number must be raised to in order to achieve another number. For instance, if we have \(x = e^y\), then \(y = \ln(x)\). The natural logarithm function, represented as \(\ln(x)\), is widely used in calculus, where 'e' is the base of the natural logarithm and is approximately equal to 2.71828.

  • In our exercise, \(\ln\) depicts the natural logarithm of an expression.
  • It鈥檚 used to simplify complex exponentials back to a linear form for easier interpretation and calculation.
  • Logarithms help in solving equations where the unknown variable is an exponent.
Exponential functions
Exponential functions are central to understanding growth processes that change multiplicatively over time. An exponential function is expressed in the form of \(y = a \cdot e^{kx}\), where 'e' is the base of the natural logarithms. This function exhibits rapid growth or decay based on the sign and magnitude of 'k'.

  • In the exercise, the expression \(e^x\) is part of the fraction inside the logarithm.
  • It's highlighting how fractions of exponential terms can be transformed for simplification.
  • This transformation often involves the manipulation of the exponents to simplify or restate the problem in a different form, aiding in further analyses such as differentiation.
Simplifying expressions
Simplifying expressions is a critical step in calculus and mathematics in general, where the goal is to reduce complex expressions into simpler forms that are easier to work with. This often involves combining like terms, reducing fractions, or factoring.

  • In the given exercise, the simplification process involves manipulating the fraction \(\frac{e^x - 1}{e^x}\).
  • By expressing \(e^x\) as a common factor, we split the terms: \((1 - \frac{1}{e^x})\).
  • Simplifying makes the expression easier to plug into the logarithm and helps in carrying out further operations such as differentiation.

Effective simplification can transform initially intimidating problems into ones that are more intuitively solvable, which is crucial for both solving equations and optimizing functions.

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Most popular questions from this chapter

Given \(\begin{aligned} f(x) &=a x(x-1)+b, \quad x<1 \\ &=x-1, \quad 1 \leq x \leq 3 \\ &=p x^{2}+q x+2, \quad x>3 \end{aligned}\) Find the constants \(a, b, p\) and \(q\) so that \(f(x)\) is differentiable at \(x=1 \& x=3\). \(\left.p=\frac{1}{3}, q=-1\right\\}\)

$$ \text { Prove that } f(x)=|\ln x| \text { is continuous but not differentiable at } x=1 \text { . } $$

$$ y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) $$

$$ y=\sqrt[3]{\frac{x\left(x^{2}+1\right)}{\left(x^{2}-1\right)^{2}}} $$

$$ y=(\tan 2 x)^{\cot ^{\frac{x}{2}}} $$
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