Use the Lambert W!!! Even if you don't need it

[@nizogos]

The x=3 solution is visible,after that you could move all the terms of your equation to one side and define a function.From there you could prove that the function is monotonous through its derivatives and thus x=3 is a unique solution

[@owlsmath]

yep makes sense!

[@taterpun6211]

At least for real numbers. Lambert W function has infinite branches like logz

[@owlsmath]

@@taterpun6211 right always infinite number of complex solutions

[@MikeMagTech]

Nice job. That was slightly headachy, but not too bad ; )

[@owlsmath]

Yep! More headaches than guess and check for sure

[@slavinojunepri7648]

Excellent

[@owlsmath]

thanks :)

[@holyshit922]

x=3 is a solution but is it the only one

[@HJ28_398]

I don't know the proper way to do it, but just doing some Desmosing, I've found that around 3.44983 - 7.89905i and 3.44983 + 7.89905i are two of these such answers, you are correct 👏👏.

[@Cyrusislikeawsome]

I guess proving it with the W function kinda helps that because you can see via the graph

[@kirbo722]

Wow. :0

[@owlsmath]

😁

[@reaperskyfall6691]

After seeing this my brain 🧠: please pull out calculator 100ms

[@owlsmath]

Is that 100 milliseconds? Your brain reacted fast!

[@reaperskyfall6691]

@@owlsmath no sir I mean cal-c type 100ms

[@owlsmath]

Oh I see!

[@maxvangulik1988]

2^x=11-x
1=(11-x)2^x
2^11•ln(2)=(11-x)ln(2)e^(11-x)ln(2)
(11-x)ln(2)=ln(256)
11-x=log_2(256)=8
x=11-8=3

[@prollysine]

2^11*ln2=ln2*(x-11)*e^((x-11)*ln2) , 2^3*2^8*ln2=ln2*(x-11)*e^((x-11)*ln2) , 8*ln2*e^(8*ln2)=ln2*(x-11)*e^((x-11)*ln2) ,
8*ln2=ln2*(x-11) , 8=x-11 , x=3 , test , 2^3+3=11 ,
W( 2^11*ln2)=~ 5.545175444 , ln2*(11-x)=~ 5.545175444 , x= -(5.545175444/ln2)+11 , x=3 ,