smeag0l:
No. Forward what you sent earlier using that email you put here.

Done

I won't show you privately. The solution you sent me for question 1 is a dubious solution and even at that, it isn't still correct. You can post it here if you want corrections so that others can learn from it.

Mrshape:

Show me my error please

I'm still waiting for the message you forwarded. I hope you sent to the email I provided in my first post?

Mathscum:

Done

smeag0l:
I won't show you privately. The solution you sent me for question 1 is a dubious solution and even at that, it isn't still correct. You can post it here if you want corrections so that others can learn from it.

I will post it here.
But I doubt you understand Lambert W
Because if you do you will not even recognize interaction which give approximate solution and graphical methods as the solution to that question

Thank God another person also solved using Lambert W
Let's see what he did too.
I post mine here

smeag0l:
No. Forward what you sent earlier using that email you put here.

Seen it?

I should be the one teaching you about Lambert W functions. I tried using this to solve that question later in the years and didnt still arrive at the beautiful answer I got using the Newton-Raphson method. The solution x=2 is just one root of the equation 2^(2x)-8x=0. The other one is 0.155. Besides, The last two or the lines of your Lambert W solution clearly shows it is dubious.(

Mrshape:

I will post it here.
But I doubt you understand Lambert W
Because if you do you will not even recognize interaction which give approximate solution and graphical methods as the solution to that question

Thank God another person also solved using Lambert W
Let's see what he did too.
I post mine here

smeag0l:
I won't show you privately. The solution you sent me for question 1 is a dubious solution and even at that, it isn't still correct. You can post it here if you want corrections so that others can learn from it.

From the point x=-An(-(ln2)/2))/ln2
That's can be classified as 100% the answers instead of going any further.
You can fake this solution anywhere no one can fault it.
That was a screen shot from nuaraland I have solved it even before you post the question.
And the other shape if you are not away it my shape
Check my previous thread you see all of them

smeag0l:
I should be the one teaching you about Lambert W functions. I tried using this to solve that question later in the years and didnt still arrive at the beautiful answer I got using the Newton-Raphson method. The solution x=2 is just one root of the equation 2^(2x)-8x=0. The other one is 0.155. Besides, The last two or the lines of your Lambert W solution clearly shows it is dubious.(

0.155 is not close to the answer.
Substitute my answer and yours into the equation and see which will give you better result

smeag0l:
I'm still waiting for the message you forwarded. I hope you sent to the email I provided in my first post?

I have forward it
It is a piece of paper

smeag0l:
I'm still waiting for the message you forwarded. I hope you sent to the email I provided in my first post?

Or should I send it here.?

I was typing a response(one meant for only mathematics intellectuals) to your earlier message on how flawed your workings were until I saw this so I had to stop. Please check my previous message after the deadline passed. This question has two solutions i..e x=2 or x =0.155.Substitute both numbers into the equation and see what you have. I can provide you a full working of these solutions according to Newton-Rahpson method if you want to learn. I will even show you calculations on how i arrived at my test values . In your solution, (-1/ln2)/(-2ln2) suddenly gives 2 instead of 1/2ln4. I am now thinking you must have copied this from somewhere online.

Mrshape:

0.155 is not close to the answer.
Substitute my answer and yours into the equation and see which will give you better result

smeag0l:
I should be the one teaching you about Lambert W functions. I tried using this to solve that question later in the years and didnt still arrive at the beautiful answer I got using the Newton-Raphson method. The solution x=2 is just one root of the equation 2^(2x)-8x=0. The other one is 0.155. Besides, The last two or the lines of your Lambert W solution clearly shows it is dubious.(

Choose for yourself which is more accurate

smeag0l:
I was typing a response(one meant for only mathematics intellectuals) to your earlier message on how flawed your workings were until I saw this so I had to stop. Please check my previous message after the deadline passed. This question has two solutions i..e x=2 or x =0.155.Substitute both numbers into the equation and see what you have. I can provide you a full working of these solutions according to Newton-Rahpson method if you want to learn. I will even show you calculations on how i arrived at my test values . In your solution, (-1/ln2)/(-2ln2) suddenly gives 2 instead of 1/2ln4. I am now thinking you must have copied this from somewhere online.

Have you check on mine?

smeag0l:
I was typing a response(one meant for only mathematics intellectuals) to your earlier message on how flawed your workings were until I saw this so I had to stop. Please check my previous message after the deadline passed. This question has two solutions i..e x=2 or x =0.155.Substitute both numbers into the equation and see what you have. I can provide you a full working of these solutions according to Newton-Rahpson method if you want to learn. I will even show you calculations on how i arrived at my test values . In your solution, (-1/ln2)/(-2ln2) suddenly gives 2 instead of 1/2ln4. I am now thinking you must have copied this from somewhere online.

Oh my fault there I downloaded the wrong solution I made two solutions that day

Yours is also dubious. You people usually do not complete the proof to the Lambert W function. I need to do a lecture on that for a fee on this forum.

Mathscum:

Have you check on mine?

So why were you arguing?The problem is that you guys don't know how to complete the solution using the Lambert W function. The Lambert W function will give you the two numbers x=2 and x=0.154953466 when carefully applied.

Mrshape:


Oh my fault there I downloaded the wrong solution I made two solutions that day

smeag0l:
So why were you arguing?The problem is that you guys don't know how to complete the solution using the Lambert W function. The Lambert W function will give you the two numbers x=2 and x=0.154953466 when carefully applied.

Sir i want to ask if logarithm is not applicable in solving question one. Because I used it and got 2 as answer.

There's a dubious part of the application of logarithm or indices to the solution which a lot of you guys included in your solution simply because when you insert x=2 into the question it solves. I saw so many submissions including these kind of solutions and I'm usually disappointed because I think that if I can spot these kind of simple errors after 10 years of graduation from the university and most of the people that turn in these answers can't it then shows the deteriorating nature of the education Nigerian universities offer these days. By the way, I never studied mathematics, I studied engineering and I still know all these.

matex97:
Sir i want to ask if logarithm is not applicable in solving question one. Because I used it and got 2 as answer.

smeag0l:
So why were you arguing?The problem is that you guys don't know how to complete the solution using the Lambert W function. The Lambert W function will give you the two numbers x=2 and x=0.154953466 when carefully applied.

Alright.
But the answer we usually leave as a function or use programing to get the numerical value.

smeag0l:
Yours is also dubious. You people usually do not complete the proof to the Lambert W function. I need to do a lecture on that for a fee on this forum.

When you do the lecture mention me.
I need to be part if it

smeag0l:
Yours is also dubious. You people usually do not complete the proof to the Lambert W function. I need to do a lecture on that for a fee on this forum.

for a fee?
Please help us with the lecture sooner so you don't forget

smeag0l:
I have contacted all 5 winners(Plus one winner I chose for myself for making a beautiful attempt to answer the first question). I am waiting for them to send their details to be credited after which I'll make another post with their monikers and screenshot of the timestamps showing wen they sent in their answers(for credibility purposes). After that, I'll ask them to post their solutions here for people to see.

Sir , did u check mine
toroabasietim@yahoo.com

smeag0l:
I have contacted all 5 winners(Plus one winner I chose for myself for making a beautiful attempt to answer the first question). I am waiting for them to send their details to be credited after which I'll make another post with their monikers and screenshot of the timestamps showing wen they sent in their answers(for credibility purposes). After that, I'll ask them to post their solutions here for people to see.

What if someone sent the correct answer before one of the guys you chose, will you still credit them? I was thinking the screenshot of the winners email with time should be posted first for transparency before crediting them??

Check my number 2,
abiolaolalekan216@yahoo.com

smeag0l:
I was typing a response(one meant for only mathematics intellectuals) to your earlier message on how flawed your workings were until I saw this so I had to stop. Please check my previous message after the deadline passed. This question has two solutions i..e x=2 or x =0.155.Substitute both numbers into the equation and see what you have. I can provide you a full working of these solutions according to Newton-Rahpson method if you want to learn. I will even show you calculations on how i arrived at my test values . In your solution, (-1/ln2)/(-2ln2) suddenly gives 2 instead of 1/2ln4. I am now thinking you must have copied this from somewhere online.

My bad. Had to draw and squeeze-in my graph in my worksheet and as a result, missed the 0.155 as it wasn't visible enough.
I don't know if you saw my mail though sent in at 2:43 early hours of yesterday. I asked but didn't get any reply there.
I know our people here no slow sha, but I just wonder if you saw it.

Your graph isnt correct. The result is a parabolic graph with a y intercept of 1.

DesChyko:


My bad. Had to draw and squeeze-in my graph in my worksheet and as a result, missed the 0.155 as it wasn't visible enough.
I don't know if you saw my mail though sent in at 2:43 early hours of yesterday. I asked but didn't get any reply there.
I know our people here no slow sha, but I just wonder if you saw it.

I had mentioned this earlier in one of my posts.

Prinss:


What if someone sent the correct answer before one of the guys you chose, will you still credit them? I was thinking the screenshot of the winners email with time should be posted first for transparency before crediting them??

You cant leave the answer as a function because that is actually not the answer. You have to evaluate it with a program like wolfram alpha. The Lambert W function usually has two solutions for all negative values greater than -1/e. The result is usually a product logarithm and if any of you had left the answer in the form a a product logarithm you would have gotten it. The last four lines of your solution were incorrect while mathslord's solution had the last two lines of his own solution as incorrect.

Mrshape:

Alright.
But the answer we usually leave as a function or use programing to get the numerical value.

smeag0l:
You cant leave the answer as a function because that is actually not the answer. You have to evaluate it with a program like wolfram alpha. The Lambert W function usually has two solutions for all negative values greater than -1/e. The result is usually a product logarithm and if any of you had left the answer in the form a a product logarithm you would have gotten it. The last four lines of your solution were incorrect while mathslord's solution had the last two lines of his own solution as incorrect.

With what you wrote here, it means it cannot be evaluated ordinarily without the use of a computer program. In that case, graphical method remains the easiest way to solve it.

smeag0l:
You cant leave the answer as a function because that is actually not the answer. You have to evaluate it with a program like wolfram alpha. The Lambert W function usually has two solutions for all negative values greater than -1/e. The result is usually a product logarithm and if any of you had left the answer in the form a a product logarithm you would have gotten it. The last four lines of your solution were incorrect while mathslord's solution had the last two lines of his own solution as incorrect.

That's exactly what I did in my solution I evaluated with wolfram alpha it gave me one value.
I am surprised when you say my solution is dubious.

I didnt look at it because it was sent in after the challenge had closed.

Loris576:

Sir , did u check mine
toroabasietim@yahoo.com

smeag0l:
You cant leave the answer as a function because that is actually not the answer. You have to evaluate it with a program like wolfram alpha. The Lambert W function usually has two solutions for all negative values greater than -1/e. The result is usually a product logarithm and if any of you had left the answer in the form a a product logarithm you would have gotten it. The last four lines of your solution were incorrect while mathslord's solution had the last two lines of his own solution as incorrect.

I've not heard of all this wolfram alpha and lambert W function before.
I thought the beauty of mathematics is that you can use any method as far as you arrive at the final answer??
If am in a competition, I only need to look at the number one question to know that x is equal to 2.