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Chapter 3: Problem 39

Solve the exponential equation algebraically. Approximate the result to three decimal places. $$e^{2 x}-4 e^{x}-5=0$$

Short Answer

Expert verified
The solution to the equation \(e^{2 x}-4 e^{x}-5=0\) is \(x = \ln(5)\), approximate to three decimal places, \(x = 1.609\)

Step by step solution

01

Reformulation of the exponential equation

View the term \( e^{x} \) as a single entity, for example \( y \). This means that \( e^{2x} \) can also be written as \( y^2 \). Therefore, the equation \( e^{2x} - 4e^x - 5 = 0 \) can be rewritten into the quadratic form of \(y^2 - 4y - 5 = 0\) where \( y = e^x \)
02

Solving the quadratic equation

The quadratic equation can be factored into \((y - 5)(y + 1) = 0\). Setting each factor equal to zero gives \(y - 5 = 0 \Rightarrow y = 5\) and \(y + 1 = 0 \Rightarrow y = -1\)
03

Solving for x

Solving the equation \( e^x = y \), where \( y = 5 \) and \( y = -1 \), we get \(x = \ln(5)\) and \( x = \ln(-1) \). However, note that \( e^x \) != -1 for any real number \( x \). Therefore, the only real solution we get is \(x = \ln(5) \) which is approximately 1.609
04

Checking the solution using the original equation

Substitute \(x = \ln(5)\) into the original equation \( e^{2x} - 4e^x - 5 = 0 \). If both sides of the equation are equal, then this confirms that the solution is correct

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Most popular questions from this chapter

Use a graphing utility to graph the function. Be sure to use an appropriate viewing window. \(f(x)=\ln x+8\)

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