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Chapter 3: Problem 60

Solve the exponential equation algebraically. Approximate the result to three decimal places. $$e^{2 x}-5 e^{x}+6=0$$

Short Answer

Expert verified
The solutions to the exponential equation are \(x = 1.099\) and \(x = 0.693\) to three decimal places.

Step by step solution

01

Setup The Quadratic Equation

Let's first rewrite the given equation in terms of exponential form. Given \(e^{2x} -5e^{x} +6 = 0\), we can let \(y = e^{x}\), re-writing the equation as a quadratic form \(y^2 - 5y + 6 = 0\).
02

Solve for \(y\) - Applying the quadratic formula

The quadratic formula is given by \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substituting \(a = 1, b = -5, c = 6\) we get \(y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4*1*6}}{2*1}\), which simplifies to \(y = \frac{5 \pm \sqrt{25 - 24}}{2}\). Hence the solutions for y are \(y_1 = 3\) and \(y_2 = 2\).
03

Find the value of \(x\)

Recall that we defined \(y = e^{x}\), so we can set our solutions for \(y\) equal to \(e^{x}\) and solve for \(x\): \(e^{x} = 3\) and \(e^{x} = 2\). Therefore, \(x = \ln(3)\) and \(x = \ln(2)\). To obtain a numerical result, we compute these values to three decimal places: \(x_1 = 1.099\) and \(x_2 = 0.693\).

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