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

1. ## 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.

2. 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.

$cb^n + kb^m = 0$

$cb^n = -kb^m$

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

$cb^{n - m} = -k$

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

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

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

-Dan

3. Thanks, that has helped a lot.

4. 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:

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

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

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

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

$b =\sqrt[n-m]{ -\frac{k}{c}}$

5. 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.

6. 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 $kb^n + cb^m - D$ 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