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 tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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