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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 \( x \):\( x = \frac{\ln(3)}{2} \).

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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.

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Most popular questions from this chapter

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