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Chapter 3: Problem 29

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

Short Answer

Expert verified
The solutions to the exponential equation are \(x = 2\) and \(x = -1\).

Step by step solution

01

Compare Exponents

Since the bases of the equation, \( e^{x} = e^{x^{2}-2} \) are the same (which is \(e\) in this case) and are equal to each other, we can equate the exponents. Therefore, the equation simplifies to \( x = x^{2} - 2 \).
02

Rearrange the Equation

Next, rearrange the equation to one side to set the equation to zero, which is the standard format for solving such equations. To do that, subtract \(x\) from both the sides of the equation to get \( x^{2} - x - 2 = 0 \).
03

Solve the Quadratic Equation

To find the value of \( x \), we can either factor out the equation or use the quadratic formula. But since the equation factorizes easily, we can use the factoring method. The factored form of the equation is \((x-2)(x+1)=0\). Setting each factor equal to zero gives the solutions \( x = 2 \) and \(x = -1\).
04

Evaluate and Check the Solutions

In this case, both solutions obtained are valid, because both satisfy the original equation \( e^{x} = e^{x^{2}-2}\). When \( x = 2 \) the equation becomes \( e^{2} = e^{4-2} \) and when \(x = -1\) the equation becomes \( e^{-1} = e^{1-2}\), and both equations are true since \( e^{2} = e^{2} \) and \( e^{-1} = e^{-1} \). Hence, the two solutions are valid.
05

Approximate the Result to Three Decimal Places

There is no need for approximation in this particular case as the obtained solutions for \(x\) are already in their simplest form.

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