A Logarithmic Exponential Equation with Euler

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[@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

[@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.

[@jackyzhang588]

Use Lambert W function will solve this question rapidly.

[@mr.d8747]

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

[@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.

[@nikko2505]

Used Lambert W-function

[@FisicTrapella]

Me too

[@georgesdermesropian1565]

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

[@kassuskassus6263]

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

[@SweetSorrow777]

Lol. Why? That's boring.😅

[@phill3986]

😊😊😊👍👍👍

[@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

[@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.

[@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