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Math Help - Help with this equation please: cb^n + kb^m = 0

  1. #1
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    Help with this equation please: cb^n + kb^m = 0

    Hey, I dont know whether you can do this or not, cant remember , but is it possible to make b the subject of this equation (like, b = ....), if so could someone explain how to do this, thanks.

    cb^n + kb^m = 0

    where c, k, n, m are constants

    Any help would be much appreicated on whether you can solve this or not.
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  2. #2
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    Quote Originally Posted by FireSoul View Post
    Hey, I dont know whether you can do this or not, cant remember , but is it possible to make b the subject of this equation (like, b = ....), if so could someone explain how to do this, thanks.

    cb^n + kb^m = 0

    where c, k, n, m are constants

    Any help would be much appreicated on whether you can solve this or not.
    I will presume that n > m, but you can easily adapt the solution if not.

    cb^n + kb^m = 0

    cb^n = -kb^m

    c\frac{b^n}{b^m} = -k

    cb^{n - m} = -k

    b^{n - m} = -\frac{k}{c}

    b = \left ( -\frac{k}{c} \right ) ^{1/(n - m)}

    Of course, whether the solution actually exists depends on the values of the constants.

    -Dan
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    Thanks, that has helped a lot.
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  4. #4
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    Quote Originally Posted by FireSoul View Post
    Hey, I dont know whether you can do this or not, cant remember , but is it possible to make b the subject of this equation (like, b = ....), if so could someone explain how to do this, thanks.

    cb^n + kb^m = 0

    where c, k, n, m are constants

    ...
    Hello,

    I assume that c is not equal zero and n and m are not equal:

    c \cdot b^n + k \cdot b^m = 0\ , m \neq n

    c \cdot b^n = -k \cdot b^m Divide both sides of the equation by c and b^m

    \frac{b^n}{b^m} = -\frac{k}{c}

    b^{n-m} = -\frac{k}{c}

    b =\sqrt[n-m]{ -\frac{k}{c}}
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  5. #5
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    Is it possible to make b the subject of this equation as well?

    kb^n + cb^m = D

    Couldnt find anything on it through google, so any help would be appreciated. Thanks.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by FireSoul View Post
    Is it possible to make b the subject of this equation as well?

    kb^n + cb^m = D

    Couldnt find anything on it through google, so any help would be appreciated. Thanks.
    I highly doubt it, unless the constants are such that kb^n + cb^m - D can be factored somehow.

    For example we CAN solve:
    b^2 + 3b = -2

    or for a more complicated example:
    2b^{2/3} - 5b^{1/3} = 3

    -Dan
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