Recent explorer of the lambert function and a question on here sparked in interest in me to do the following tutorial: Solving a linear(t)=exponential(t) for the variable t algebraically using the lambert function.

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- freckles

- schrodinger

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- freckles

\[ \text{ How to solve something of the form (for } t \text{):} \\ at+b=c(d)^t \\ \text{ using the Lambert Function } \\
\text{ Left side is a linear expression whereas } \\ \text{ the right side is a exponential expression}\\ \text{ anyways I will solve this one in general } \\
\text{ First step: factor out the left hand side } \\ \text{ we want to factor out the coefficient of } t \\ at+b=a(t+\frac{b}{a}) \\ \text{ so now we have the equation as } a(t+\frac{b}{a})=c(d)^t \\
\text{ Second step: divide both sides by } c \\ \frac{a}{c}(t+\frac{b}{a})=d^{t} \\
\text{ Third step: sub} t+\frac{b}{a} \text{ as } u \\ \text{ so if } u=t+\frac{b}{a} \text{ then } t=u-\frac{b}{a} \\ \text{ So our equation looks like } \frac{a}{c}u=d^{u-\frac{b}{a}} \\
\text{ Fourth step: multiply both sides by } d^{\frac{b}{a}} \\ \text{ this gives us } \\ \frac{a}{c}u d^{\frac{b}{a}}=d^u \\
\text{ Fifth step: multiply } u^{-1} \text{ on both sides } \\ \frac{a}{c} d^{\frac{b}{a}}=u^{-1}d^u \\
\text{ Sixth step: rewrite } d^u \text{ as } (\frac{1}{d})^{-u} \\ \text{ so we have} \\ \frac{a}{c} d^{\frac{b}{a}}=u^{-1}(\frac{1}{d})^{-u}
\text{ Seventh step: “flip” both sides } \\ \frac{c}{a} d^{-\frac{b}{a}}=u(\frac{1}{d})^u \\
\text{ Eight step: recall } e^{\ln(\frac{1}{d})}=\frac{1}{d} \\ \frac{c}{a} d^{-\frac{b}{a}}=u(e^{\ln(\frac{1}{d}})^u
\text{ Ninth step: Multiply both sides by} \ln(\frac{1}{d}) \\ \ln(\frac{1}{d}) \frac{c}{a} d^{-\frac{b}{a}}=\ln(\frac{1}{d}) u(e^{\ln(\frac{1}{d})})^u
\text{ Tenth step: Take } W( ) \text{ (this is the Lambert function } \\ \text{ of both sides } \\ \text{ Note: recall } W(be^b)=b \\ W(\ln(\frac{1}{d}) \frac{c}{a} d^{-\frac{b}{a}})=W(\ln(\frac{1}{d}) u(e^{\ln(\frac{1}{d})})^u) \\ \text{ which can be written as } W(\ln(\frac{1}{d}) \frac{c}{a} d^{-\frac{b}{a}})=\ln(\frac{1}{d})u
\text{ Eleventh step: Divide both sides by } \\ \ln(\frac{1}{d}) \text{ aka} -\ln(d) \\
\frac{-1}{\ln(d)}W(\ln(\frac{1}{d}) \frac{c}{a}d^{-\frac{b}{a}})=u \\
\text{ Twelfth step: recall the sub we made earlier } u= t+\frac{b}{a} \\
\text{ So the solution is } t=\frac{-1}{\ln(d)}W(\ln(\frac{1}{d}) \frac{c}{a}d^{-\frac{b}{a}})-\frac{b}{a}\]

- freckles

And I should probably place some restrictions on the constant values in my problem.

- freckles

\[a \neq 0 \\ d \in (0,1) \cup (1,\infty)\]

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## More answers

- freckles

what if a is 0 though?
\[b=c(d)^t\]
this equation should be tons simpler
\[\frac{b}{c}=d^t \\ \ln(\frac{b}{c})=\ln(d^t) \\ \ln(\frac{b}{c})=t \ln(d) \\ \frac{1}{\ln(d)} \ln(\frac{b}{c})=t \]
As we see the Lambert function isn't needed for a=0.

- freckles

What if d=1?
Well you no longer have an exponential expression on the right hand side. This equation is tons easier than the last.
\[at+b=c \\ at=c-b \\ t=\frac{c-b}{a}, a \neq 0\]

- freckles

Anyways I might play more with Lambert and see if I can post anything else at a later time.

- freckles

one more restriction for my first problem c is not 0

- rational

Thats really a nice general method to express/solve using product log function! Did you come up with that method ? xD

- freckles

Yeah. I had to do a few playings...
Though I did kind of get an idea for the sub here http://en.wikipedia.org/wiki/Lambert_W_function .

- rational

@Kainui taught me this some time back... I want to review but looks its gonna take time as I don't seem to remember much

- freckles

Yeah I kind of remember him I think mentioning something about Lambert

- rational

\[\large x^{x^{x^\vdots}}=?\]

- freckles

I didn't try to get it back then. But the question asked today may be kind of interested though I don't think it was what the asker actually intended. http://openstudy.com/users/freckles#/updates/557a14dce4b07028ea5fb518

- freckles

i'm playing with the thingy you have there
I already know the answer because I look at math world or whatever it is called
but I want to see if I can derive it

- rational

\[\large x^{x^{x^\vdots}}= t \implies x^t = t \]
thats in general form that you derived earlier

- rational

guess i can start with Fifth step straight

- freckles

\[x^t=t \\ x=e^{\ln(x)} \\ e^{\ln(x) \cdot t}= t \\ t^{-1} e^{\ln(x) \cdot t}=1 \\ t^{-1} e^{\ln(x) \cdot t \cdot (-1)(-1)}=1 \\ t^{-1}e^{\ln(x^{-1} )\cdot t(-1)}=1 \\ (te^{\ln(x^{-1})t})^{-1}=1 \\ te^{\ln(x^{-1} )\cdot t}=1 \\ \ln(x^{-1})te^{\ln(t^{-1}) \cdot t} =\ln(x^{-1}) \\ \ln(x^{-1} )t=W(\ln(x^{-1}))\]
so
\[t=\frac{W(\ln(x^{-1}))}{\ln(x^{-1})}\]

- freckles

but maybe I made an algebraic mistake

- freckles

oh no actually I think that looks good

- freckles

the string of exponents of x can be replaced by t
and that equation holds because both t's represent an infinite exponent string of x's
if you know what I mean?

- rational

Nice! wolf gives the exact same expression
http://www.wolframalpha.com/input/?i=solve+x%3Dt%5Ex

- rational

Yes its just another form of infinite sequence... it has to converge in the first place for our analysis to make any sense

- freckles

not bad
and not much of a difference from what I was doing earlier with the other equation

- freckles

oh wait it is the same thing lol

- freckles

that is an exponent=linear

- freckles

i'm dumb :p

- freckles

I think I got threw off by the power tower

- rational

Haha! im looking at discrete lambert w function, its more interesting than the continuous version xD

- freckles

I will have to look at that tomorrow
I must go tonight
thanks for the fun power tower

- rational

thanks again for sharing the general method.. gnite!

- Kainui

Here are two great sites I found when learning about the product log:
http://www.had2know.com/academics/lambert-w-function-calculator.html
https://luckytoilet.wordpress.com/tag/product-log/
There are some exercises on the first link, which I think are fun to try to work out without cheating.

- freckles

\[W=xe^{-W } \\ W'=e^{-W}-W'xe^{-W} \\ W'+W'xe^{-W}=e^{-W } \\ W'(1+xe^{-W})=e^{-W} \\ W'=\frac{e^{-W}}{1+xe^{-W}} \\ W'=\frac{1}{e^{W}+x} \\ \text{ but since } W=xe^{-W} \implies e^W=\frac{x}{W} \\ \text{ then } W'=\frac{1}{\frac{x}{W}+x}=\frac{W}{x+Wx}\]
ok just wanted to do that one thing myself
:p

- Kainui

Haha also might as well find the indefinite integral too ;P

- freckles

\[\int\limits W(x) dx \\ \text{ Let } y=W(x) \\ \text{ then } dy=W'(x) dx \\ W'(x)=\frac{W(x)}{x+W(x)x}=\frac{y}{x+yx} \\ \text{ but if } y=W(x) \text{ then } W^{-1}(y)=x \\ \text{ and } W^{-1}(y)=ye^y \\ \text{ so } x=ye^{y} \\ W'(x)=\frac{y}{ye^y+y^2e^y} \\ \text{ back \to } dy=W'(x) dx \\ \frac{dy}{W'(x) }=dx \\ \frac{ye^y+y^2e^y}{y} dy=dx \\ (e^y+ye^y) dy=dx \\ \int\limits W(x) dx=\int\limits y (e^y+ye^y)dy\]

- freckles

\[\int\limits W(x) dx=\int\limits (ye^y+y^2e^y) dy \\ ye^y-e^y+y^2e^y-2ye^y+2e^y+C \text{ by IBP}\]

- freckles

\[=-ye^y+e^y+y^2e^y+C\]

- freckles

now we need to write back in terms of x

- freckles

\[x=ye^y \\ \int\limits W(x) dx=x+e^y+yx+C\]
we have to figure how to take care of the exp(y) and the y thingy there
\[y=W(x)\]
\[\int\limits W(x) dx=x+e^{W(x)}+W(x) \cdot x+C\]

- freckles

hmm their answer looks a little different than mine

- freckles

oh I forgot to bring down my negative factor for that one term

- freckles

\[\text{ So it should read } \\ \int\limits W(x) dx=-x+e^{W(x)}+W(x) \cdot x+C\]
and I could have replaced exp(W) with x/W like they did
but don't really see a reason to

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