How do you solve #x+2=e^(x) #?

2 Answers
Oct 11, 2016

Answer:

Use Newton's Method

#x = 1.146193# and #x = -1.84141#

Explanation:

You cannot solve the equation using algebraic methods. For this type of equation, I use a numerical analysis technique called Newton's Method.

Here is a reference to Newton's method

Let #f(x) = e^x - x - 2 = 0#

#f'(x) = e^x - 1#

You start with a guess for #x_0# and then do the following computation to move closer to the solution:

#x_(n+1) = x_n - f(x_n)/(f'(x_n))#

You do computation, feeding each step back into the equation, until the number that you get doesn't change from the previous number.

Because Newton's Method is computationally intensive, I use an Excel Spreadsheet.

  1. Open an Excel Spreadsheet

Into cell A1 enter your guess for #x_0#. I entered 1 into cell A1.

Into cell A2 enter the following expression:

=A1 - (EXP(A1) - A1 - 2)/(EXP(A1) - 1)

Copy the contents of cell A2 into the clipboard and then paste it into cell A3 through A10.

You will see that the number quickly converges on #x = 1.146193#

Edit: After reading a very nice comment from Shell. I decided to find the second root by changing the value of cell A1 from 1 to -1. The spreadsheet quickly converges on the value #x = -1.84141#

Oct 11, 2016

Answer:

This question cannot be solved algebraically. Graphing gives #x=-1.841# and #x=1.146#.

Explanation:

The left side of the equation #x+2# is algebraic.

The right side of the equation #e^x# is transcendental (it can't be expressed as a polynomial e.g. exponentials, logs, trig functions).

This equation cannot be solved algebraically but it can be solved graphically.

To solve, plot both #color(red)(y=x+2)# and #color(blue)(y=e^x)# in a graphing utility or graphing calculator. The solutions are the #x# coordinates of the intersections.

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