/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 98 Find the value of \(b\) for whic... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 98

Find the value of \(b\) for which $$1+e^{b}+e^{2 b}+e^{3 b}+\cdots=9$$

Short Answer

Expert verified
The value of \(b\) is \(b = \ln\left(\frac{8}{9}\right)\).

Step by step solution

01

Recognize the Series

The given series is a geometric series: \[1 + e^{b} + e^{2b} + e^{3b} + \dots\] The first term \(a\) is 1, and the common ratio \(r\) is \(e^b\). This expression refers to the infinite geometric series sum formula.
02

Apply the Infinite Series Sum Formula

To find the sum of the infinite geometric series, use the sum formula: \[S = \frac{a}{1 - r}\]For this series: \[S = \frac{1}{1 - e^b}\]We know this sum equals 9.
03

Set up the Equation

We set the infinite series sum equal to 9:\[\frac{1}{1 - e^b} = 9\] This equation can be solved for \(b\).
04

Solve for \(e^b\)

First, take the reciprocal of both sides to simplify the equation:\[1 - e^b = \frac{1}{9}\]Then, rearrange the equation to express \(e^b\):\[e^b = 1 - \frac{1}{9} = \frac{8}{9}\]
05

Solve for \(b\)

Now, to find \(b\), take the natural logarithm of both sides to solve for \(b\):\[b = \ln\left(\frac{8}{9}\right)\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Geometric Series Sum Formula
An infinite geometric series is a sequence where each term is a constant multiple of the previous one. In the problem provided, the series is:
  • First term: 1
  • Common ratio: \(e^b\)
The sum of an infinite geometric series can be calculated with the formula:\[S = \frac{a}{1 - r}\]where \(a\) is the first term and \(r\) is the common ratio, provided \(|r| < 1\).

In the exercise, we are given that the sum \(S\) is 9. Applying the formula:\[9 = \frac{1}{1 - e^b}\]This equation helps in setting up a relationship between the terms in the series and their sum. Solving this allows us to find the common ratio in terms of the exponent \(b\).
Solving Exponential Equations
Exponential equations involve variables in the exponents, such as \(e^b\). To solve them, we can manipulate the equation to isolate the exponential expression on one side. Using the exercise example, we started with:\[\frac{1}{1 - e^b} = 9\]By taking the reciprocal on both sides, we get:\[1 - e^b = \frac{1}{9}\]Next, rearrange to find \(e^b\):\[e^b = 1 - \frac{1}{9} = \frac{8}{9}\]This process usually involves basic algebraic techniques to navigate around the exponential form. Isolating \(e^b\) was crucial here, as it sets the stage for further solving using logarithms.
Natural Logarithm
The natural logarithm, denoted as \(\ln\), is used to solve for a variable in an exponential equation. Because logarithms are the inverse functions of exponentials, they allow us to "bring down" the exponent.

In the exercise, once we isolated \(e^b\) as \(\frac{8}{9}\), we solved for \(b\) by applying the natural logarithm:\[b = \ln\left(\frac{8}{9}\right)\]This is a straightforward step once the equation is set up correctly.
  • The natural logarithm especially simplifies calculations when dealing with the base \(e\) (approximately equal to 2.71828), which is common in continuous growth scenarios.
  • Understanding how natural logarithms work is crucial for solving exponential equations like the one in the exercise.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which of the series in Exercises 13 46 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=2}^{\infty} \frac{\sqrt{n+2}-\sqrt{n+1}}{\sqrt{n+1} \sqrt{n+2}} $$

Which of the sequences converge, and which diverge? Give reasons for your answers. $$ a_{n}=\frac{2^{n}-1}{3^{n}} $$

How many terms of the convergent series \(\sum_{n=4}^{\infty} 1 /\left(n(\ln n)^{3}\right)\) should be used to estimate its value with error at most 0.01\(?\)

Estimating Pi The English mathematician Wallis discovered the formula \begin{equation}\frac{\pi}{4}=\frac{2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdot 8 \cdot \cdots}{3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdot 7 \cdot \cdots}\end{equation} Find \(\pi\) to two decimal places with this formula.

Determine if the sequence is monotonic and if it is bounded. $$ a_{n}=\frac{2^{n} 3^{n}}{n !} $$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.