Thread: 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.

Originally Posted by FireSoul

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.













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

-Dan

Thanks, that has helped a lot.

Originally Posted by FireSoul

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:



Divide both sides of the equation by c and b^m

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.

Originally Posted by FireSoul

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 can be factored somehow.

For example we CAN solve:


or for a more complicated example:


-Dan