How Can You Solve the Equation e^x = 5-2x?

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To solve the equation e^x = 5 - 2x, traditional algebraic methods are ineffective due to the presence of x in both the exponent and as a linear term. The discussion suggests using the natural logarithm to rewrite the equation, but acknowledges that isolating x is not straightforward. A numerical solution or the Lambert W function is recommended for finding an approximate solution. Graphical methods, such as plotting the functions and identifying intercepts, can also provide a practical approach. Overall, the conversation emphasizes the need for numerical techniques to solve this type of transcendental equation.
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I am a little stuck how to solve this equation

e^x = 5-2x?

I did ln e^x = ln (5-2x)

x = ln(5-2x) / ln e
but iam not sure how to bring the other x around to the side with the x to solve the equation?
 
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You can't with the "usual" functions. Since you have x both in the exponent and not, you would have to do a numerical solution or use "Lambert's W function", the inverse function to f(x)= xex.
 
how would you go about doing a numerical solution?
 
Why not try it graphically and then do some approximating to find the intercept, e.g. decimal search. This would work if you did not need a highly accurate solution, e.g. in terms of Pi.

The Bob (2004 ©)
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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