Thread: Exponential algebra

Howdy folks. First time poster here

Have a bit of a problem and can't seem to figure the correct way to solve it

I'm looking for Ka

0.93Ka - 1.969 = -e^(-2Ka)

Any ideas in how to solve it? I was planning on taking LN of both sides, but that still leaves me with LN of Ka on one side with Ka on the other side of the equation, which is practically the same.

just for curiosity,is this problem has anything to do with chemistry ?

Originally Posted by Raoh

just for curiosity,is this problem has anything to do with chemistry ?

Sort of. It's to do with the calculation of the reaeration factor for a lake.

Any ideas?

I thought it has something to do with acids,'cause there's this interesting relation $\displaystyle PKa=-\log Ka$.
(Acidity constant).

Hello FreemasonFurbie

Welcome to Math Help Forum!

Originally Posted by FreemasonFurbie

Howdy folks. First time poster here

Have a bit of a problem and can't seem to figure the correct way to solve it

I'm looking for Ka

0.93Ka - 1.969 = -e^(-2Ka)

Any ideas in how to solve it? I was planning on taking LN of both sides, but that still leaves me with LN of Ka on one side with Ka on the other side of the equation, which is practically the same.

You won't be able to find an algebraic solution. Your best bet is a numerical method. With a spreadsheet, I've just found a couple of solutions: one negative and one positive. To 4 d.p. they are:

$\displaystyle K_a=-0.4315, \;+2.1011$

I hope that helps.

Grandad

Originally Posted by Grandad

Hello FreemasonFurbie

Welcome to Math Help Forum!You won't be able to find an algebraic solution. Your best bet is a numerical method. With a spreadsheet, I've just found a couple of solutions: one negative and one positive. To 4 d.p. they are:

$\displaystyle K_a=-0.4315, \;+2.1011$

I hope that helps.

Grandad

Thanks! Would you mind explaining how you did it in Excel? I'd be interested in seeing how you did it.

I wasn't aware that it had such a function. I was going to use Matlab, but alas, I forgot to install it on my computer after my last restore.

Hello FreemasonFurbie

Originally Posted by FreemasonFurbie

Thanks! Would you mind explaining how you did it in Excel? I'd be interested in seeing how you did it.

I wasn't aware that it had such a function. I was going to use Matlab, but alas, I forgot to install it on my computer after my last restore.

Nothing very subtle - trial and error, really. I just set up the function 0.93x+exp(-2x) and tried some likely values around x=0. Then refined them when I got a general feel for the behaviour of the function.

Without a spreadsheet I should use Newton's Method - but to be honest it's loads quicker just to stick some numbers into Excel! (You'd never think I was a real mathematician, would you?)

Spreadsheet attached (for what it's worth!)

Grandad

You chancer!!!

haha.....

Thanks Grandad. The help is much appreciated!

Originally Posted by FreemasonFurbie

Howdy folks. First time poster here

Have a bit of a problem and can't seem to figure the correct way to solve it

I'm looking for Ka

0.93Ka - 1.969 = -e^(-2Ka)

Any ideas in how to solve it? I was planning on taking LN of both sides, but that still leaves me with LN of Ka on one side with Ka on the other side of the equation, which is practically the same.

Solve numerically the solution in (1,5) is: $\displaystyle Ka\approx 2.101$

Can also be solved in terms of Lambert's W function:

$\displaystyle Ka=\frac{1}{2}\;W\left[-\frac{2}{a}\; e^{-2b/a}\right]+\frac{b}{a}$

CB