How to invert this exponential function to solve for x:  $y = a \exp(bx) + c \exp(dx)$? Cheers.
So if I don't make sense, I have a value for $y$, I need to know what $x$ is.
$$y = a \exp(bx) + c \exp(dx)$$
$a =       12.85$,
$b =    0.001857$,
$c =      -54.24$,
$d =    -0.05316$
 A: Write $\displaystyle b = B/M$ and $\displaystyle d = D/M$, where $\displaystyle M \gt 0$ and $\displaystyle B,D$ are integers (I am assuming $\displaystyle b,d$ are rational).
Set $\displaystyle z = e^{x/M}$
We get
$\displaystyle y = az^B + c z^D$
This is a polynomial in $\displaystyle z$. You could try using polynomial root finding methods now.
A: I doubt there is an analytic solution.
With those values of $a,b,c,d$, it appears that $y$ is smoothly increasing and bijective function of $x$ so given any value of $y$ you should be able to find a very close approximation $x$ by standard numerical methods.  
For example, if $y=0$ then $x \approx 26.175\ldots$, if $y=100$ then $x \approx 1104.9\ldots$, if $y=-100$ then $x \approx -13.7278\ldots$.
A: You have $y=ae^{bx} + ce^{dx}$ with $a, b$ positive and $c, d$ negative. So this means:


*

*if $x$ is large and positive, then $y \approx ae^{bx}$

*if $x$ is large and negative, then $y \approx ce^{dx}$
So for $x$ large and positive, $x \approx (1/b) \log (y/a)$; for $x$ large and negative, $x \approx (1/d) \log (y/c)$ -- these come from solving the above approximations for $x$. If you need to go further I'd say start with these approximations and then use something like Newton's method.
