/*! 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 33 Solve the equation. $$e^{4 x}+... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 33

Solve the equation. $$e^{4 x}+4 e^{2 x}-21=0$$

Short Answer

Expert verified
\( x = \frac{\ln(3)}{2} \)

Step by step solution

01

Substitute

Let's introduce a substitution to make the equation simpler.Let \( y = e^{2x} \). Then the equation becomes \( y^2 + 4y - 21 = 0 \). This substitution helps us form a quadratic equation.
02

Solve the Quadratic Equation

We now need to solve the quadratic equation \( y^2 + 4y - 21 = 0 \).We can factor this equation as follows:\( (y + 7)(y - 3) = 0 \).Thus, the solutions for \( y \) are \( y = -7 \) and \( y = 3 \).
03

Determine the Valid Solution

Since \( y = e^{2x} \), it must be positive, so \( y = -7 \) is not a valid solution. Thus, we only consider \( y = 3 \).
04

Substitute Back to x

Now substitute back to express in terms of \( x \):Since \( y = 3 \), we have \( e^{2x} = 3 \).Take the natural logarithm of both sides:\( 2x = \ln(3) \).Solve for \( x \):\( x = \frac{\ln(3)}{2} \).

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.

Quadratic Substitution
Quadratic substitution is a powerful technique used to simplify complicated equations, particularly exponential equations, into more manageable forms. When dealing with exponential equations like \( e^{4x} + 4e^{2x} - 21 = 0 \), substituting \( y = e^{2x} \) enables us to reformulate the problem into a quadratic equation. This is because \( e^{4x} \) becomes \( y^2 \), allowing us to rewrite the original equation as \( y^2 + 4y - 21 = 0 \).
  • This method simplifies solving because quadratic equations have established solution methods.
  • Using substitution converts complex exponential terms into easy-to-handle polynomial terms.
Once the equation is simplified, it is easier to factor, solve for \( y \), and eventually work back to find \( x \). Remember, the key is to choose a substitution that directly leads to a recognizable form.
Natural Logarithms
Natural logarithms are crucial when solving for variables in the exponent, particularly in exponential equations like \( e^{2x} = 3 \). The natural logarithm, denoted as \( \ln \), is the inverse operation of exponentiation with the base \( e \). To solve \( e^{2x} = 3 \), we take the natural logarithm of both sides: \[ \ln(e^{2x}) = \ln(3) \].Using the property \( \ln(e^a) = a \), this simplifies to \( 2x = \ln(3) \). This step is essential in turning the problem from the exponential form into a linear one that is much easier to solve.
  • Natural logarithms help 'bring down' exponents for effective processing.
  • They are integral in equations where the variable is an exponent, easing the path to isolating it.
Once simplified, you can solve for \( x \) by dividing both sides by the appropriate constant.
Factoring Quadratic Equations
Factoring is a fundamental method for solving quadratic equations which appear frequently via substitution methods. For the equation \( y^2 + 4y - 21 = 0 \), factoring involves finding two numbers that multiply to \(-21\) and add up to \(4\). These numbers are \(7\) and \(-3\), so we can factor the quadratic as \((y + 7)(y - 3) = 0\).
  • Factoring can be thought of as 'unzipping' the polynomial into components that are easier to solve.
  • It provides direct solutions to \( y \) when set each factor equal to zero, e.g. \( y + 7 = 0 \) or \( y - 3 = 0 \).
By factoring, we find potential solutions \( y = -7 \) and \( y = 3 \). However, in the context of the original problem where \( y = e^{2x} \), only positive solutions are physically meaningful, so \( y = -7 \) is discarded. Always re-evaluate solutions in the context of the full mathematical situation at hand.

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

These exercises use the population growth model. The population of the world was 5.7 billion in 1995 and the observed relative growth rate was 2% per year. (a) By what year will the population have doubled? (b) By what year will the population have tripled?

These exercises use the population growth model. Culture starts with 8600 bacteria. After one hour the count is 10,000. (a) Find a function that models the number of bacteria \(n(t)\) after \(t\) hours. (b) Find the number of bacteria after 2 hours. (c) After how many hours will the number of bacteria double?

These exercises use the radioactive decay model. The mass \(m(t)\) remaining after \(t\) days from a 40 -g sample of thorium- 234 is given by $$m(t)=40 e^{-0.0277 t}$$ (a) How much of the sample will remain after 60 days? (b) After how long will only 10 g of the sample remain? (c) Find the half-life of thorium-234.

How long will it take for an investment of \(\$ 1000\) to double in value if the interest rate is \(8.5 \%\) per year, compounded continuously?

A learning curve is a graph of a function \(P(t)\) that measures the performance of someone learning a skill as a function of the training time \(t .\) At first, the rate of learning is rapid. Then, as performance increases and approaches a maximal value \(M,\) the rate of learning decreases. It has been found that the function $$P(t)=M-C e^{-k t}$$ where \(k\) and \(C\) are positive constants and \(C
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

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.