I need to solve this equation for 'B'. I am having trouble rearranging the equation to achieve a 'B=' answer. Can someone help me rearrange this equation to solve for B?

0=((DCRMg) - (DCRMgAl)) ∗ e^(-B∗((ρMg)+(ρAl)))+ ((DCRMgAl )- (DCRAl ))∗e^(-2∗B∗(ρMg) )+((DCRAl) - (DCRMg)) e^(-B∗((ρMg)+(ρMgAl)))

Hey LanieH.

Can you show us what you have tried? You might want to start off by moving the 2nd and later terms to the LHS and then taking logarithms.

Hi Chiro.
I have not even attempted to solve this. I do not know how to solve an equation like this for B. Are you able to give it a try? Do you know of anyone who may be able to solve this?

Is Al one variable or it is A*L

Originally Posted by LanieH
I need to solve this equation for 'B'. I am having trouble rearranging the equation to achieve a 'B=' answer. Can someone help me rearrange this equation to solve for B?

0=((DCRMg) - (DCRMgAl)) ∗ e^(-B∗((ρMg)+(ρAl)))+ ((DCRMgAl )- (DCRAl ))∗e^(-2∗B∗(ρMg) )+((DCRAl) - (DCRMg)) e^(-B∗((ρMg)+(ρMgAl)))
D|C|R=0

This eq simplifies to:

$m g(1- a l) * e^{-b \rho ( m g+ a l)}+ a l( m g-1)*e^{-2b \rho m g}+( a l- m g) *e^{-b \rho m g( 1+ a l)}=0$

If you have numerical values of the parameters then b can be evaluated! otherwise I can't see a closed form solution