# Solving an equation where the unknown appears in under the exponent and as a constant

I encountered a problem trying to find a solution to the following equation (the problem 9.45 from the wonderful textbook on Applied Calculus by Hoffman et.al (2013, p. 719)):

$$A(t) = \frac{3}{k} (1 - e^{-kt})$$

According to the exercise, we know that $A(1) = 2.3$ implying that

$$2.3 = \frac{3}{k} (1 - e^{-k})$$

Here I don't know what to do. I played with the equation back and forth trying to multiply and divide both sides of it with all sorts of things but it did not bring me much. Of course, I can multiply both sides by $k$

$$2.3 k = 3 (1 - e^{-k})$$

And it looks like $k = 0$, which cannot be correct as the whole equation does not make sense in terms of the context the problem is given (it is about the content of a drug in patience's bloodstream, therefore there has to be a real-number solution.)

Wolfram Alpha suggests that the answer is $k = 0.557214$

Could you suggest me an analytical solution to the equation?

• This equation is a "transcendental equation", so there is likely no simple formula for $k$ in terms of so-called "elementary functions". But there are plenty of ways of approximating the solution, which is what Wolfram Alpha has done. – Morgan Sherman Mar 21 '17 at 22:26
• I asked a similar question before and was given a helpful answer here math.stackexchange.com/questions/1735552/… – WaveX Mar 21 '17 at 22:28

The "analytical" answer is $$k = \frac{30}{23} + W\left(-\frac{30}{23} e^{-30/23}\right)$$ where $W$ is the Lambert W function.
Just a note: it cannot be $k=0$ not only because it does not make sense, but because you would be dividing by 0 in the first equation.