# Rearranging formula with exponents, for the subject of:?

Hi there,

I'm trying solve a fairly long equation for a specific variable. I can do most of it but have begun running into problems with small parts of it.

For example, the equation contains terms like 2e^-(c/d) and -icos(x)e^-((f+c)/d), and I'm not sure how to simplify and solve these for d. Before, I was taking the natural log of both terms separately, but then realized log(A*B) is not equal to (log(a) +log(b))

So just to take a small part of the equation I am solving, how would I solve the following for d?

1-e^(-g/d) = 2e^(-c/d) -icos(x)e^-((f+c)/d)

Thanks a lot!

I'm trying solve a fairly long equation for a specific variable. I can do most of it but have begun running into problems with small parts of it.

For example, the equation contains terms like 2e^-(c/d) and -icos(x)e^-((f+c)/d), and I'm not sure how to simplify and solve these for d. Before, I was taking the natural log of both terms separately, but then realized log(A*B) is not equal to (log(a) +log(b))

So just to take a small part of the equation I am solving, how would I solve the following for d?

1-e^(-g/d) = 2e^(-c/d) -icos(x)e^-((f+c)/d)

Thanks a lot!

## 2 Answers

- log(A·B) = log(A) + log(B)00
- This looks to me the sort of equation that cannot be solved algebraically, especially if some of the variables are complex--as it seems they need to be. If you let z = e^(-1/d), it seems you effectively have

.. 1 - z^g = 2z^c - icos(x)z^(f+c)

which you want to solve for z.

If the variables are real, the cos(x) term must be zero (as there are no imaginaries anywhere else), but you still have

.. 1 = z^g + 2z^c

Unless g and c enjoy a useful relationship (one is double or triple the other, for example), this "polynomial" seems unlikely to be solved by algebraic means.

If some of the variables are complex, it may be useful to separate the equation into real and imaginary parts.00

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