Logarithms: Weight W = Fat Free Mass (FFM) + Fat Mass (FM), formula FFM(t) = kln(FM(t)/m)

[Jon Trainer]

It's 30 years since I have done anything like this and, to be honest, I think my algebra brain has withered!
Anyway, here goes.
In the world of weight loss the following relationship is often used:

Weight W = Fat Free Mass (FFM) + Fat Mass (FM)

Some clever people have derived a relationship between FFM and FM that generally holds true, which is this
FFM(t) = kln(FM(t)/m)

Where k and m are constants and ln is the natural logarithm. Here FFM and FM are functions of time but, since this relationship holds true for all t under consideration, I am ignoring the time dependency for the purpose of solving a bit of algebra.

I'm looking for an expression for FFM in terms of W. Clearly I can substitute the first expression into the second, like so
FFM = kln((W-FFM)/m)

, which leaves just two variables W and FFM. So, in principle, I should be able to rearrange this to derive an expression for FFM in terms of W.
Can I figure it out? No. I've tried rearranging it and taking exponents etc, but I can't figure it out. Perhaps it can't be done without a Taylor series expansion or something of that sort.

Can anyone put me out of my misery?
Many thanks

 

[tkhunny]

It's 30 years since I have done anything like this and, to be honest, I think my algebra brain has withered!
Anyway, here goes.
In the world of weight loss the following relationship is often used:

Weight W = Fat Free Mass (FFM) + Fat Mass (FM)

Some clever people have derived a relationship between FFM and FM that generally holds true, which is this
FFM(t) = kln(FM(t)/m)

Where k and m are constants and ln is the natural logarithm. Here FFM and FM are functions of time but, since this relationship holds true for all t under consideration, I am ignoring the time dependency for the purpose of solving a bit of algebra.

I'm looking for an expression for FFM in terms of W. Clearly I can substitute the first expression into the second, like so
FFM = kln((W-FFM)/m)

, which leaves just two variables W and FFM. So, in principle, I should be able to rearrange this to derive an expression for FFM in terms of W.
Can I figure it out? No. I've tried rearranging it and taking exponents etc, but I can't figure it out. Perhaps it can't be done without a Taylor series expansion or something of that sort.

Can anyone put me out of my misery?
Many thanks

You have tripped over a very common difficulty. It cannot be solved with a nice, algebraic expression. You have noticed that anything you do to remove one FFM from the logarithm entraps the other FFM and gets you nowhere. This does not mean you cannot find a solution for some set of parameters. There are various numerical and graphical solutions available.

 

[Jon Trainer]

Thanks tkhunny - that's useful to know as it will stop me endlessly scribbling alternative forms of the expression!
My challenge is that I have hacked together a spreadsheet and I have 400 rows of data on which to apply the hoped-for formula. I know that on the few rows I tried a solution exists because the Excel solver found it each time. So I think my problem has moved from being algebraic to figuring out a way to get Excel to find the solution for all 400 rows without having to use the actual solver screen.
Thank you for taking the time to answer.

 

[tkhunny]

Thanks tkhunny - that's useful to know as it will stop me endlessly scribbling alternative forms of the expression!
My challenge is that I have hacked together a spreadsheet and I have 400 rows of data on which to apply the hoped-for formula. I know that on the few rows I tried a solution exists because the Excel solver found it each time. So I think my problem has moved from being algebraic to figuring out a way to get Excel to find the solution for all 400 rows without having to use the actual solver screen.
Thank you for taking the time to answer.

Macros, My Friend. Macros.

 

[Jon Trainer]

Indeed. Thanks.