Been too many years since algebra...

[Greg Truby]

OK, I'm stumped...

If I have:

m = ba<SUP>ec</SUP> + d


and I need to solve for e then I do:

• m - d = ba<SUP>ec</SUP>
• (m-d)/b = a<SUP>ec</SUP>
• log<SUB>a</SUB>((m-d)/b) = ec
• (log<SUB>a</SUB>((m-d)/b))/c = e

But what if I have:

k = e × m

where m is as above? i.e.

k = e × (ba<SUP>ec</SUP> + d)

Now how the dickens can I solve for e, i.e. figure out the value of e if all the other variables' values are known?

 

[Andrew Fergus]

Hi Greg

Unless I'm mistaken, can't you start with :

k / e = (ba<SUP>ec</SUP> + d)

and then proceed as you did before?

i.e.
(k / e) - d = ba<SUP>ec </SUP>etc.

Andrew

 

[nichclark]

Andrew that wouldn't leave e on it's own Greg needs e=..
Greg be careful with the notation your using log base e times e = 1. but I don't think the e your using is e.

For the answer I'm not sure either but i'm sure there is some wizz out there who can help.

 

[Andrew Fergus]

Sorry - my bad. I didn't see the e repeated......

 

[TestPoint]

k = e × (ba<SUP>ec</SUP> + d)


is the "x" which is after the "e" a multiplication sign, or is it another variable?

 

[RossMcColl]

k = me

where e = ((loga((m-d)/b))/c)

so you end up with the two following:

k/m = ((loga((m-d)/b))/c)
m = baec + d

=

k/baec + d = ((loga((m-d)/b))/c)

and solve it out from there

 

[Greg Truby]

Andrew, yes it's the repetion both as a coeffient and as an exponent that's got me tripped up.

Ross, perhaps I'm missing your point? I know the values of k, a, b, c & d But I don't know e or m. I tried going down that path already and if there's a way through, I don't see it.

Nichclark - my bad, even though my original formulae have that variable as e, I should have forseen someone confusing it with e as in e=2.718282... or ln(e) = 1.

Testpoint, yes it's multiplication's × not a variable x; again my apology for the lack of clarity.

Let me re-type this using the classic x for the unknown and dot multiplier.

How does one solve the following for x?

• k = x · (ba<SUP>xc</SUP> + d)

where k, a, b, c & d are known values?

 

[hatman]

I got as far as

• (m-d)/(ba^c)=ea^e

And from what I can get out of the log laws, I don't think you can get the e by itself from here...

 

[Oaktree]

It's with good reason that you're struggling with this, Greg.... I don't think it can be done.

Consider that x + ln(x) = y cannot be solved for x, save for iterative processes for a known y.

This is what Goal Seek is for

 

[Greg Truby]

hatman, oaktree,

Thanks to both of you for taking a look and at least making this old boy feel a bit better about his quite rusty algebra skills.

...This is what Goal Seek is for...

Yeah, that was Plan B. I just wanted to make sure that there wasn't a direct route.

Thanks a bunch to everyone that spent a moment of his time on this one.