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Chapter 3: Problem 31

Solve the exponential equation algebraically. Approximate the result to three decimal places. $$e^{x^{2}-3}=e^{x-2}$$

Short Answer

Expert verified
The solutions to the given exponential equation are \(x_{1}=1.618\) and \(x_{2}=-0.618\), approximated to three decimal places.

Step by step solution

01

Set the exponents equal to each other

Given that the base numbers are equal (as they're both e), this allows us to set the two exponents equal to each other. Thus, the exponential equation \(e^{x^{2}-3}=e^{x-2}\) becomes a quadratic equation: \(x^{2}-3 = x-2\).
02

Rearrange the equation

Rearrange the equation \(x^{2}-3 = x-2\) to make it standard form for quadratic equations by bringing all terms to the left side. In this case, the equation becomes \(x^{2} - x - 1 = 0\).
03

Use the quadratic formula to solve for \(x\)

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\) to solve for \(x\). For the equation \(x^{2} - x - 1 = 0\), \(a = 1\), \(b = -1\), and \(c = -1\). Substituting these values into the quadratic formula gives the solutions \(x_{1}, x_{2}= \frac{1 \pm \sqrt{5}}{2}\).
04

Approximate the result

As per the exercise's request: approximate the result to three decimal places. Thus, \(x_{1} = 1.618\) and \(x_{2}= -0.618\).

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