How do I solve $e^{ax}-e^{bx}=c$ for $x$?

Question

I have the following general question:

How do I solve $e^{ax}-e^{bx}=c$ for $x$?

The constants $a$, $b$ and $c$ are real numbers. For context, this is the final form of a longer equation that I have simplified.

The actual, more specific, equation I'm trying to solve applies to a radioactive generator of molybdenum-99/technetium-99m in my nuclear medicine department. We want to know what is its exact calibration time.

The equation is:

$A_T = A_M \lambda_T (e^{-\lambda_Mt}-e^{-\lambda_Tt}) / (\lambda_T-\lambda_M)$

$A_T$ is the activity of Tc-99m at elution time monday morning = about 3 Curies

$A_M$ is the activity of Mo-99 at calibration time on sunday = 6 Curies

$\lambda_T$ is the decay constant for for Tc-99m = 0.1152

$\lambda_M$ is the decay constant for for Mo-99 = 0.01051

t is the time between calibration and elution. It's what we want to determine.

The goal is to do the first elution each monday, mesure the activity of Tc-99m obtained (it's usualy between 3 and 3.5 Ci) and find out at what time on sunday the generator was calibrated for its 6 Ci activity of Mo-99. So all the variables are fixed except fot t and $A_T$.

Answer 1

I'll simply reiterate what others have said: There is no general way to solve equations of the form you have. There are a handful of values of $a, b$ for which a solution to $\;e^{ax} - e^{bx} = c\;$ can always be solved: e.g., if $a, b \in \{1, 2, 3, 4\}$, you're in luck. (Of course, if $c = 0$, you're also in luck.)

There are numerical methods for solving such an equation for given values of $a, b, c$. If you have only the general form you posted, I'm afraid there's no all-encompassing solution. But if you have particular values of $a, b, c$ in mind, feel free to edit the post and let us have a look at those values.

Answer 2

Declaring $y=e^x$ gives

$$y^a-y^b=c$$

For large $a$ and $b$ this equation is not solvable: it's either a polynomial of high degree or some other complicated beast. Numerics are the way to go here.

Answer 3

As others have said, you are in for a numeric solution, but it can be made simpler. It looks like you measure $t$ in hours, so it is in the range $10-30$, which means $ax$ is in the range $-.1$ to $-.3$ and $bx$ is in the range $-1$ to $-3$. I would rewrite the equation as $x=\frac 1a \log (c+e^{bx})$ and iterate, starting with $x_0=20$. It should converge quickly.

Answer 4

Write it as $(e^x)^a$ and the same for the second one too.

Now check if $c$ is positive or negative. If +ve, take $(e^x)^b$ common. Now the left thing $((e^x)^a)-1$ is not divisible by $e^b$ anymore. So divide $c$ by $e^b$ (if $c$ has a factor). So the no. of times you can divide it by it is $x$.

Just put $x$ afterwards and surely check.

Just visualize it and then do it. You can use it to solve equations like $(2^m)-(2^n)=56$.

Actually someone gave me the equation above (the one with powers of $2$) and then asked me to find $m$ and $n$. I divided $56$ by $2$ as many times as i can until i reached $7$ ($3$ times). So $n$ is $3$ and $m$ is $6$.

Surely try this too!

Answer 5

After you edited, your equation is:

$$ e^{-\lambda_M t} - e^{-\lambda_T t} = \frac{A_T(\lambda_T−\lambda_M)}{A_M\lambda_T}. $$ Now you want to get a reasonably accurate $t$ by numerics to approximate the true time.

Here is my suggestion: due to the derivative of the left hand side decays very fast to zero, common root finding algorithms like Newton's method (or any other root finding method belonging in the class of Householder's methods) may not converge very well.

You may wanna try the old school Bisection method.