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# C3 Help needed Tweet

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1. C3 Help needed

I've done the first part but how would I do part b?
2. Re: C3 Help needed
(Original post by 0range)

I've done the first part but how would I do part b?
3. Re: C3 Help needed
(Original post by Zii)
What would you do from there?

I mean I can get from the form they want it in to x+lnx = 0
would that be sufficient?
4. Re: C3 Help needed
(Original post by 0range)
What would you do from there?

I mean I can get from the form they want it in to x+lnx = 0
would that be sufficient?
From what I wrote, I am not sure how you are unable to get the desired result. Add x to 2x. This gives you 3x. Now divide both sides by 3.
5. Re: C3 Help needed
(Original post by 0range)

I've done the first part but how would I do part b?
Zii has already mentioned a method. But another way could have been to convert , then reverse the steps to convert .
6. Re: C3 Help needed
x + lnx = 0
x = -lnx
3x = -lnx + 2x
x = (2x - lnx) / 3

7. Re: C3 Help needed
(Original post by Zii)
From what I wrote, I am not sure how you are unable to get the desired result. Add x to 2x. This gives you 3x. Now divide both sides by 3.
Right sorry heads going a bit blank atm

(Original post by raheem94)
Zii has already mentioned a method. But another way could have been to convert , then reverse the steps to convert .
Yh I think this might be easier thanks
8. Re: C3 Help needed
(Original post by raheem94)
Zii has already mentioned a method. But another way could have been to convert , then reverse the steps to convert .
(Original post by 0range)
Yh I think this might be easier thanks
You know, my method is secretly this method in disguise.