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Chapter 4: Problem 124

Solve the equation. \(e^{2 x}-6 e^{x}+4=0\)

Short Answer

Expert verified
The solutions are \( x = \ln(3 + \sqrt{5}) \) and \( x = \ln(3 - \sqrt{5}) \).

Step by step solution

01

Introduce Substitution

To simplify the equation, introduce a substitution. Let \( u = e^{x} \). Then, \( e^{2x} \) can be rewritten as \( u^2 \). So, the equation becomes \( u^2 - 6u + 4 = 0 \).
02

Solve the Quadratic Equation

Solve the quadratic equation \( u^2 - 6u + 4 = 0 \) using the quadratic formula: \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). For our equation, \( a = 1 \), \( b = -6 \), and \( c = 4 \).
03

Apply the Quadratic Formula

Calculate the discriminant: \( b^2 - 4ac = (-6)^2 - 4 \cdot 1 \cdot 4 = 36 - 16 = 20 \). Now apply the quadratic formula: \( u = \frac{6 \pm \sqrt{20}}{2 \cdot 1} = \frac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5} \). Hence, \( u = 3 + \sqrt{5} \) and \( u = 3 - \sqrt{5} \).
04

Reverse the Substitution

Substitute back \( u = e^x \). This gives two equations: \( e^x = 3 + \sqrt{5} \) and \( e^x = 3 - \sqrt{5} \).
05

Solve for x

Take the natural logarithm of both sides to solve for \( x \). For \( e^x = 3 + \sqrt{5} \), \( x = \ln(3 + \sqrt{5}) \). For \( e^x = 3 - \sqrt{5} \), \( x = \ln(3 - \sqrt{5}) \). Check if \( 3 - \sqrt{5} > 0 \).
06

Verify the Solutions

Verify if \( 3 - \sqrt{5} > 0 \). Since \( \sqrt{5} \approx 2.236 \), \( 3 - \sqrt{5} \approx 0.764 \), which is positive. Thus, both solutions are valid.

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Key Concepts

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

Substitution Method
The substitution method is a powerful tool for solving complex equations by reducing them to simpler forms. In our example, we simplify the exponential equation by letting \( u = e^x \). This transforms the original equation \( e^{2x} - 6e^x + 4 = 0 \) into a quadratic equation: \( u^2 - 6u + 4 = 0 \).
  • First, identify a substitution that makes the equation easier to handle.
  • Replace the variables accordingly, resulting in a more familiar equation form.
  • Solve this simpler equation before reversing the substitution.
This approach allows us to use methods designed for simpler equations, like the quadratic formula, instead of directly tackling the original challenging problem.
Quadratic Formula
The quadratic formula is essential for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). In our exercise, we need to solve \( u^2 - 6u + 4 = 0 \) using the formula: \[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -6 \), and \( c = 4 \). The steps are:
  • Identify coefficients \( a \), \( b \), and \( c \).
  • Compute the discriminant \( b^2 - 4ac \).
  • Substitute these values into the quadratic formula to find solutions for \( u \).
Applying this method gives us the roots: \ u = 3 + \sqrt{5} \ \ u = 3 - \sqrt{5} \ These solutions help us move forward to the next step of solving the original equation.
Natural Logarithm
Taking the natural logarithm is vital for solving equations where the unknown is in the exponent. After finding the roots \( u = 3 + \sqrt{5} \) and \( u = 3 - \sqrt{5} \), we substitute back to get two exponential equations: \ e^x = 3 + \sqrt{5} \ \ e^x = 3 - \sqrt{5} \ To solve for \( x \), we take the natural logarithm (\( \ln \)) of both sides:
  • For \( e^x = 3 + \sqrt{5} \), \( x = \ln(3 + \sqrt{5}) \).
  • For \( e^x = 3 - \sqrt{5} \), \( x = \ln(3 - \sqrt{5}) \).
The natural logarithm function reverses the exponentiation, letting us isolate \( x \) and solve the equations.
Discriminant
The discriminant is a crucial concept in the quadratic formula. It helps determine the nature and number of roots. For the quadratic equation \( u^2 - 6u + 4 = 0 \), the discriminant \( \Delta \) is calculated as \( b^2 - 4ac \): \[ \Delta = (-6)^2 - 4 \cdot 1 \cdot 4 = 36 - 16 = 20 \]
  • If \( \Delta > 0 \), the equation has two distinct real roots.
  • If \( \Delta = 0 \), there is one real double root.
  • If \( \Delta < 0 \), the equation has no real roots, only complex ones.
Here, \( \Delta = 20 > 0 \) indicates two distinct real roots. This is consistent with our solutions: \ u = 3 + \sqrt{5} \ \ u = 3 - \sqrt{5} \ Understanding the discriminant aids in predicting the solvability and nature of the equation's roots.

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Most popular questions from this chapter

Solve the equation. Write the solution set with the exact values given in terms of common or natural logarithms. Also give approximate solutions to 4 decimal places. \(21,000=63,000 e^{-0.2 t}\)

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Solve the equation. Write the solution set with the exact solutions. Also give approximate solutions to 4 decimal places if necessary. \(\ln x+\ln (x-4)=\ln (3 x-10)\)

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