## Changing the subject of this equation...

I am having difficulty making x the subject of the following formula.

y = [a.e^(b.x)] + [c.e^(d.x)]

I thought the first step would be to take the natural log of both sides of the equation:

ln(y) = ln(a)+b.x+ln(c)+d.x

But this does not work, even though the following is correct:

y = a.e^x
ln(y) = ln(a) + x

I am a little stuck as to what to try next!

Dan

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 Quote by dannybeckett I am having difficulty making x the subject of the following formula. y = [a.e^(b.x)] + [c.e^(d.x)] I thought the first step would be to take the natural log of both sides of the equation: ln(y) = ln(a)+b.x+ln(c)+d.x But this does not work, even though the following is correct: y = a.e^x ln(y) = ln(a) + x I am a little stuck as to what to try next! Dan
The problem is, you can't make x the subject! At least it's not expressible in terms of finitely many elementary functions we commonly use, such as logs, powers, trig etc.
 I had a feeling this was going to be the outcome... damnit!!

Mentor

## Changing the subject of this equation...

 Quote by dannybeckett I am having difficulty making x the subject of the following formula. y = [a.e^(b.x)] + [c.e^(d.x)] I thought the first step would be to take the natural log of both sides of the equation: ln(y) = ln(a)+b.x+ln(c)+d.x
The reason this doesn't work is that ln(A + B) ≠ ln(A) + ln(B). You cannot take the log of a sum and get the sum of the logs.
 I see, thankyou. So there is no way at all to make x the subject in this case?

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 Quote by dannybeckett I see, thankyou. So there is no way at all to make x the subject in this case?
There is, but it'd be in terms of an infinite sum, which is even less useful than just finding a numerical approximation to x.