Question

Here's a fun problem I just came up with using Lambert's product log W(x):

Answer 1

\[\LARGE W(e^x*b^b )=b*b^b\]

Solve for b. Good luck! Medal to the winner. =P

@ganeshie8 @dan815 @ikram002p @iambatman

Answer 2

\(\large e^x*b^b = b*b^b e^{b*b^b}\)

\(\large b^b[e^x - b e^{b*b^b}] = 0\)

Answer 3

You're on the right track, but you haven't solved for b yet @rational

Answer 4

What is W ? a constant ?

Answer 5

@HatcrewS nope, it's the product log function, check it out:

https://luckytoilet.wordpress.com/tag/product-log/
http://www.had2know.com/academics/lambert-w-function-calculator.html

Answer 6

then what is the value of W

Answer 7

it has no value, just like "ln" has no value. It's a function.

Answer 8

why is the the exponent of the left hand side equal to the whole left hand side.

Answer 9

makes no sense @@

Answer 10

What do you mean? I don't follow

Answer 11

it says that when u remove an X from the exponent, the power doesnt change

Answer 12

where, show me the equation you're talking about

Answer 13

At any rate, continuing where rational left off we can conclude from \[\LARGE b^b(e^x−be^{bb^b})=0\] this means either \[\LARGE b^b=0\] or \[\LARGE e^x−be^{bb^b}=0\] The first one is pretty obviously \[\LARGE b \ln b = \ln0=-\infty\] but that's not really the answer we're looking for, since it's not a number. So let's try to continue with the second one to get the real equation for b in terms of x.

Answer 14

wolfram doesn't like either of the equations

Answer 15

Hahaha well I'm sorry you can't cheat.

Answer 16

http://www.wolframalpha.com/input/?i=solve+y%2C++e%5Ex-++y*e%5E%28y*y%5Ey%29%3D0

Answer 17

\(\large e^x - b e^{b*b^b}= 0 \implies x = \ln b + b*b^b\)