Let's Solve A Homemade Exponential

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[@EricA.Ephemetherson]

Very nice and not easy. I tried to use logs but I lost it after a few moves. I did not know that you could manipulate the polynomial as the exponent to get the Lambert. In this case it was adding 1 to (x^2+2x) in e. But what was done to the RHS to equate that? Multiplying 1 by e? If that's the case then it was kind of a constriction on numbers. If it does not fit, force it.

[@honestadministrator]

exp (( ( x + 1) ^2 - 1) /2) = - 1/( x + 1) exp ( (z^2 - 1) /2) = - 1/z Has a trivial solution at x + 1 = z = -1 This implies x = - 2

[@Archimedes_Notes]

Clearly -2 is the only solution. The range values set tne value of x to be -2. Archimides decomposition wiill work as well.Equations everyday.yeah..

[@jaimeduncan6167]

Very nice problem and solution.

[@SyberMath]

Thank you 🙂

[@barakathaider6333]

👍

[@giuseppeimbimbo1555]

We should have expected that after squaring both sides we could have introduced extraneous solutions. That's what happened with x=0!

[@jaimeduncan6167]

He mentioned it.

[@ashmitsarkar6192]

Cool Loved it

[@vladimirrodriguez6382]

Well done

[@sandeeshsewanagala8829]

It's nice 👍👍👍

[@scottleung9587]

Nice!

[@yamikira6512]

Can you multiply by (x+1) and integrate both sides? LHS is f'e^f

[@giuseppemalaguti435]

??? Non si può fare

[@yamikira6512]

​@@giuseppemalaguti435 (x+1)e^(x+x^2/2)=-1 Integrating from -1 to some X we get: e^(x+x^2/2)= -(x+1) But we know that LHS is also equal to -1/(x+1), thus: -(x+1) = -(1/x+1) (x+1)^2 = 1 => x = -2 or x = 0 I wonder if it's a coincidence

[@giuseppemalaguti435]

x=-√W(e)-1=-2

[@ziplock007]

I am always confused how solving for solutions and then you check to find some 'don't work' Any approach in these "Lambert's" problems to only find working solutions?

[@元兒醬]

It's wrong from about 4 minutes

[@jyl123]

what’s wrong

[@元兒醬]

Sorry I was wrong