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Simple Powers/Logarithm Finding solutions? Watch

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    Hey, well I've always struggled for one reason or the other with powers/logarithms, I've never truly understood all the rules involved.

    Essentially I've boiled a long arse problem down to this, for which I need to find the solutions:

    2^n + n  2^n = 4^n

    I'm told the solutions are 0 and 1, but truly have no idea how to solve it. I try logs but I get lost and don't really know what I'm doing. Help much appreciated!
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    Factorise.
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    (Original post by Mr M)
    Factorise.
    2^n + n  2^n - 4^n = 0

4^n = 2^{2n} = (2^n)^2



2^n + n  2^n - (2^n)^2 = 0

2^n (1 + n + 2^n) = 0

    From there I'm a little lost, 2^n = 0 has no roots. Not sure what do with  (1 + n + 2^n) = 0 .

    Also not entirely sure how what rule I'm using to go from 4^n to 2^2n, I know it works because I've tested it, but it was more through trial and error that.
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    (Original post by AndroidLight)
    ...
    I did read that before you deleted and you correctly discovered that the problem reduces to:

    2^n=n+1

    You could have divided by 2^n as you realised that it cannot equal zero but I didn't want to encourage division without thought.

    There is no nice way of solving this now.

    The best approach is to sketch y=2^x and y=x+1 and realise there must be 2 solutions. One is immediately obvious and the second is trivial to find.

    Edit: You have some sign issues.
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    Cheers, that makes good sense. I'll make a note of the drawing graph method as that'll sure come in handy too. I got the intersections as 0 and 1, which is what I'm looking for. I still have way too much a tendency too jump into logs but hopefully that'll come with experience.

    Cheers again, much appreciated.
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    (Original post by AndroidLight)
    Cheers, that makes good sense. I'll make a note of the drawing graph method as that'll sure come in handy too. I got the intersections as 0 and 1, which is what I'm looking for. I still have way too much a tendency too jump into logs but hopefully that'll come with experience.

    Cheers again, much appreciated.
    You are welcome.

    This sort of equation cannot be solved with logs. It leads to something horrible called the Lambert W function.
 
 
 
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