Find the value of x using the Lambert W function

[chwala]

Homework Statement

I got this problem from wikipedia, the problem is on the attachment.

Solve the equation ##3^x=2x+2## using the Lambert W function.

Relevant Equations

Lambert W function

it really took me time to figure out as to how that equation in aterisk was arrived at...i just noted that both sides of the equation were multiplied by ##-\frac {1}{3}## any particular reason for that? what is the common thinking around that? can one multiply both sides by say ##\frac {1}{6}##? i am encountering this lambert for the first time and i really want to understand how it works...i would appreciate guidance leading to solution. Can one use the ti nspire to find the Lambert W value, in reference to the very last step leading to ##-0.79011## and ##1.44456##?

 

[chwala]

In the following step we have,
let##y=3^{-x-1}##
##ln y=(-x-1)ln 3##
##e^{(-x-1)ln3} = y##
therefore on substituting back we get,
##(-x-1)e^{(-x-1)ln3}=-\frac {1}{6}## ...note that we get the desired form ##we^w##
multiplying both sides by ##ln 3## yields,
→##ln 3 (-x-1)e^{(-x-1)ln3}=-\frac {1}{6}####ln 3##

this steps are clear no problem here, the next step i would like some input...

 

[chwala]

i could as well say,
##(x+1)3^{-x}=\frac {1}{2}##
on multiplying both sides by ##-\frac {1}{9}## we have,
##(-x-1)3^{-x-2}=-\frac {1}{18}##
it follows that, as in the previous example, after taking natural logs...
##(-x-1)e^{(-x-2)ln 3}=-\frac {1}{18}##
##ln 3 (-x-1)e^{(-x-2)ln 3}=-\frac {1}{18}ln3##
which is in the desired form of ##we^w##...this is perfectly understood...looks like one has to consider the powers of ##3## ie ##3^1, 3^2 ... 3^n##
what is remaining is to understand how they found the solution to the problem and secondly, if my approach would give an equivalent or rather correct solution...

i just checked this...this will not be the right approach since we do not get the desired form ##we^w##

 

[chwala]

Homework Statement:: I got this problem from wikipedia, the problem is on the attachment.

Solve the equation ##3^x=2x+2## using the Lambert W function.
Relevant Equations:: Lambert W function

View attachment 273868
it really took me time to figure out as to how that equation in aterisk was arrived at...i just noted that both sides of the equation were multiplied by ##-\frac {1}{3}## any particular reason for that? what is the common thinking around that? can one multiply both sides by say ##\frac {1}{6}##? i am encountering this lambert for the first time and i really want to understand how it works...i would appreciate guidance leading to solution.

I noted that we can only multiply in powers of ##3## just to answer my own question, we can not therefore, think of multiplying both sides by ##\frac {1}{6}##

 

[chwala]

I came up with another equation that is my own ...
consider ##4^x=6x+6##
→##(-x-1) 4^{-x}=-\frac {1}{6}##
→##(-x-1)4^{-x-1}=-\frac {1}{24}##
→##(-x-1)ln 4 e^{(-x-1)ln 4}= -\frac {ln4}{24}##...this is correct as we are getting the desired form of ##we^w##
→##(-x-1)ln 4=W_{0} [\frac {-ln4}{24}##].....a
→##ln 4(-x-1)=W_{-1} [\frac {-ln4}{24}##]....b
i am trying to demonstrate the fact that i am now conversant with the approach...i just need to know how to deal with the last part ##a## and ##b## cheers

 

[fresh_42]

Have you checked your results with WolframAlpha? They say it's ##x=-\dfrac{1}{\log 3}\left(W_n\left(-\dfrac{\log 3}{6}\right)+\log 3\right)##.

You can also use the series expansion to calculate the value, see
https://en.wikipedia.org/wiki/Lambert_W_function

 

[chwala]

Have you checked your results with WolframAlpha? They say it's ##x=-\dfrac{1}{\log 3}\left(W_n\left(-\dfrac{\log 3}{6}\right)+\log 3\right)##.

You can also use the series expansion to calculate the value, see
https://en.wikipedia.org/wiki/Lambert_W_function

Fresh is the solution shown on wolframalpha different from the solution that was shown on Wikipedia?ie on post ##1##

 

[fresh_42]

I haven't checked what you have written as it is a bit difficult to read. In any case your numbers ##18## and ##\log 4## didn't look right. Post #2 started well, but why do you want to divide by ##9##?

Let's see how I would start: If we substitute ##y=x+1## then we get ##6y=3^y##. With ##u=y \log 3## we get ##e^u## on the RHS and ##cu:=\dfrac{6}{\log 3}u## on the LHS which is almost what we are looking for: ##\dfrac{1}{u}e^u=c##. Next try ##v=\dfrac{1}{u}## and so on.

At the end, you should find one substitution for ##y## instead of three and write it down systematically.

 

[chwala]

I haven't checked what you have written as it is a bit difficult to read. In any case your numbers ##18## and ##\log 4## didn't look right. Post #2 started well, but why do you want to divide by ##9##?

Let's see how I would start: If we substitute ##y=x+1## then we get ##6y=3^y##. With ##u=y \log 3## we get ##e^u## on the RHS and ##cu:=\dfrac{6}{\log 3}u## on the LHS which is almost what we are looking for: ##\dfrac{1}{u}e^u=c##. Next try ##v=\dfrac{1}{u}## and so on.

At the end, you should find one substitution for ##y## instead of three and write it down systematically.

Maybe you didn't have time to look at my post...I was asking if it would be correct ...following the example on Wikipedia. I have though taken consideration of your input...thanks

 

[fresh_42]

Maybe you didn't have time to look at my post...I was asking if it would be correct ...following the example on Wikipedia. I have though taken consideration of your input...thanks

Post #2 looks good. But you made a mistake: it isn't ##e^{-2-x}\ln 3##. It has to be ##e^{(-2-x)\ln 3}## which makes the rest going wrong.

 

[chwala]

Typo error aaaaargh ...can I go ahead and amend that?

 

[fresh_42]

Typo error aaaaargh ...can I go ahead and amend that?

Sure. But you will get a multiple of ##x## so you will probably need a ##w=ax+b## substitution of so.

 

[chwala]

I am really not getting what you are saying. Look at the original post ##1##. My post ##2## and ##3## is simply an exact repetition of post ##1##. I wanted to know if ##3## is correct as I used a different factor ...What i simply wanted to know is how the author arrived at the solutions of ##-0.79## and ##1.44## that is the only part I would like to understand.
Post ##5## is my own created problem,...I was simply trying to follow post ##1## in order to solve it. I hope this is clear, I just made some typo errors but my working is correct.

 

[fresh_42]

I am really not getting what you are saying. Look at the original post ##1##. My post ##2## and ##3## is simply an exact repetition of post ##1##. I wanted to know if ##3## is correct as I used a different factor ...What i simply wanted to know is how the author arrived at the solutions of ##-0.79## and ##1.44## that is the only part I would like to understand.
Post ##5## is my own created problem,...I was simply trying to follow post ##1## in order to solve it. I hope this is clear, I just made some typo errors but my working is correct.

Sorry, I ignored the picture in post #1, because I thought it is from a book and I normally ignore any photos, because they are hard to read and my eyesight isn't what it was anymore.

Yes, these seem to be correct. WolframAlpha only confirmed the ##W_1## solution, so I could only take the graph to check the ##W_0## solution, but it looks o.k.

 

[chwala]

sorry, I should have indicated the source of the question. The source is Wikipedia.

 

[chwala]

i just amended my equations, kindly help me arrive to the solution for my post number ##5##

 

[fresh_42]

i just amended my equations, kindly help me arrive to the solution for my post number ##5##

I came up with another equation that is my own ...
consider ##4^x=6x+6##
→##(-x-1) 4^{-x}=-\frac {1}{6}##
→##(-x-1)4^{-x-1}=-\frac {1}{24}##
→##(-x-1)ln 4 e^{(-x-1)ln 4}= -\frac {ln4}{24}##...this is correct as we are getting the desired form of ##we^w##
→##(-x-1)ln 4=W_{0} [\frac {-ln4}{24}##].....a
→##ln 4(-x-1)=W_{-1} [\frac {-ln4}{24}##]....b
i am trying to demonstrate the fact that i am now conversant with the approach...i just need to know how to deal with the last part ##a## and ##b## cheers

It looks o.k. to me. At this point I would do one of the following:

• look at WolframAlpha for the general solutions and the graph
  https://www.wolframalpha.com/input/?i=4^x=6x+6
• draw the W-function graph on your own and solve the problem how to find x graphically
• take the series for the W-function, e.g. from Wikipedia, plug in the value -ln 4 / 24 and calculate the result as long as you get the precision that you want

 

[chwala]

Thanks I will definitely look at it...cheers goodnight

 

[chwala]

fresh thank you, ...it means that we have to use the computer software to finally calculate the solution...