A Homemade Exponential Log Equation

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[@mcwulf25]

I used method 1 but without the substitution. Easy to show the two terms on the lhs are identical and (lnx)^2 = 1.

[@maxwellarregui814]

Buenas noches Señores SyberMath. Es un buen ejercicio, El identificar la manera de iniciar a resolver un problema es clave, Gracias por las opciones, hay que tener ingenio e imaginación para aplicar la teoría que nos esforzamos en aprender. Más ejercicios de este tipo, que sea con factoriales. límites algébricos, trigonométricos en fin de todos los temas. Éxitos.

[@SyberMath]

Thank you for the kind words! 🥰

[@doctorb9264]

Excellent problem and solutions .

[@SyberMath]

@@doctorb9264 thank you!

[@SidneiMV]

x = eˡⁿˣ => xˡⁿˣ = (eˡⁿˣ)ˡⁿˣ
2(eˡⁿˣ)ˡⁿˣ = 2e => (eˡⁿˣ)ˡⁿˣ = e
(lnx)² = 1 => lnx = ± 1
*x = e ∨ x = 1/e*

[@SyberMath]

A nice way to approach it!

[@MichaelRothwell1]

​@@SyberMathHere is the 3rd method! (this is how I did it).

[@nagemkahwage5928]

Thank you! 👍🏻

[@SyberMath]

You're welcome!

[@Nothingx303]

Sir how can I send you questions???😊

[@SyberMath]

Check the form in the descriptions or email

[@Mediterranean81]

Rewrite every term in base e
2*e^ln(x)^2 = e^1+ln 2
e^(ln 2 + (ln x)^2) = e^(1+ln 2)
ln 2 +(ln x)^2 = 1+ ln 2
( ln x)^2 = 1
ln x = 1 or ln x = -1
x = e or x = 1/e

[@anestismoutafidis4575]

=> e^(lne)^2 + e^lne= 1e+1e=2e
x=e

[@Nothingx303]

προσθέστε αλάτι την επόμενη φορά

[@barberickarc3460]

Easy one today i guess;
e, 1/e as solutions

[@SweetSorrow777]

Using that kind of substitution is cheating.😂
Nicely done, btw.

[@SyberMath]

😜 hehe! Thanks

[@scottleung9587]

Nice!

[@SyberMath]

Thanks!

[@phill3986]

✌️😃✌️👍✌️😃✌️👍

[@rakenzarnsworld2]

x = e

[@broytingaravsol]

piece of cake

[@vladimirkaplun5774]

🚸

[@FrancisZerbib]

EZ