A Logarithmic Exponential Equation with Euler

🤩 Hello everyone, I'm very excited to bring you a new channel (SyberMath Shorts).
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡
/ @sybermath
/ @aplusbi
⭐ Join this channel to get access to perks:→ bit.ly/3cBgfR1
My merch → teespring.com/stores/sybermat...
Follow me → / sybermath
Subscribe → kzbin.info?sub...
⭐ Suggest → forms.gle/A5bGhTyZqYw937W58
If you need to post a picture of your solution or idea:
intent/tweet?text...
#algebra #exponential #logarithm
via @KZbin @Apple @Desmos @NotabilityApp @googledocs @canva
PLAYLISTS 🎵 :
▶ Trigonometry: • Trigonometry
▶ Algebra: • Algebra
▶ Complex Numbers: • Complex Numbers
▶ Calculus: • Calculus
▶ Geometry: • Geometry
▶ Sequences And Series: • Sequences And Series

[@mystychief]

What's so difficult? equation 1 can be written as xe^x=-1e^-1 so x=-1. That's it (Indeed, I did not prove it's the only solution).

[@mcwulf25]

And you don't even have to Lambert each side to solve it - although that would solve it too.

[@user-ei3sn7pm4k]

apply product log to both sides W(xe^x) = W(-1/e) x = W(-1/e) x = W_0(-1/e) = W_1(-1/e) = -1 note -1/e is a special value for the product log function

[@YouTube_username_not_found]

Let f be the function from R to R such that f(x) = xexp(x) . We analyse the function and find out that it hits a minimum at x = -1 equal to -1/e thus x = -1 is a solution and that the function is strictly convex so it is the only solution. That's all.

[@mr.d8747]

*By the definition of the Lambert W function, x = W(-1/e), with the principal value being W₀(-1/e) = -1.*

[@jackyzhang588]

Use Lambert W function will solve this question rapidly.

[@georgesdermesropian1565]

You can use the Lambert function to the equation. W(x.e^x)=W(_1/e) then x= w(_1/e)

[@davidsousaRJ]

I solved it in 5 seconds. xe^x = -1/e = (-1)e^(-1), therefore x = -1.

[@maxhagenauer24]

I see the W(-1/e) and the -1 solution immediately from the thumbnail.

[@ghstmn7320]

Solution: Assume f(x)=xe^x+1/e f is differentiable as a product and sum of differentiable functions f'(x)=(1+x)e^x If we solve f'(x)>=0 then x>=-1 and since f is continuous at [-1,+inf) then f is increasing at that interval (lets call it A1) therefore f' is negative at (-nf,-1] and since f is continuous at that interval (lets call it A2), f is decreasing there also we observe that f(-1)=0 for all x in A1, where f is increasing: x>=-1 => f(x)>=f(-1) => f(x)>=0 for all x in A2, where f is decreasing: x f(x)>=0 therefore f has a total minimum at x=-1 the position f the min is unique because, if there was another position where theres a min lets say c, then by fermat's theorem ( whose standards f obliges since: f is differentiable in all R (and therefore continuous in it) so it is differentiable in c (which is an element of R) and f presents a local extremum in c) then f'(c)=0. However, f' can be simplified into a first degree polynomial, which has only one root, therefore the position of the minimum is unique. Therefore, f(x)>=0 where the equality holds only for x=-1. So f(x)=0 has a unique root x=-1 and then so will the equation

[@phill3986]

😊😊😊👍👍👍

[@kassuskassus6263]

How about our friend the W Lambert function ? It takes 5 seconds to find x.😇😇😇

[@SweetSorrow777]

Lol. Why? That's boring.😅

[@BruceLee-io9by]

Anyone who recommends solving the equation with Lambert has not understood anything about the beauty of mathematics. This is a great video.

[@mcwulf25]

Must admit, W was my first thought!

[@BruceLee-io9by]

@mcwulf25 It was also my thought but I don't really like using Lambert.

[@SidneiMV]

xe^x = -1e^(-1) x = -1

[@nikko2505]

Used Lambert W-function

[@FisicTrapella]

Me too