Finding k in exponential continuous random variable distribution

Question

Find the unknown parameter $k$ given this distribution: $$ f(x)= \begin{cases} e^{kx} & 0 < x < 2\\ 0 & \text{otherwise} \end{cases}$$

This is my attempt:

$$\int_0^2 e^{kx} dx = \left[\frac{1}{k}e^{kx}\right]_0^2 \\ \frac{1}{k}e^{2k} -\frac{1}{k}=1 $$

This is where I am just stuck. How do I solve the above for $k$? I feel like I am missing something simple but I just can't see it.

Edit: I can see that I can rearrange this to equal $e^{2k} = k+1$ and then take logs giving me $k=\frac{1}{2}ln(k+1)$ but that still doesn't let me solve for $k$?

Answer 1

The total integral should indeed be $1$, but $ \int_0^2 e^{kx} dx = \left[ \frac{1}{k} e^{kx} \right]^{x=2}_{x=0} = \frac1k(e^{2k}-1) =1$ instead..

So $e^{2k} = (k+1)$ and take logs..

Answer 2

I think there is an error in the text. I suppose that the density is

$$f(x)=k e^{kx}$$

so

$$k=\frac{\log2}{2}$$

Answer 3

First note $k\ne0$, since $\int_0^2dx=2\ne1$. Let's put @tommik's idea aside for the moment. As @HennoBrandsma noted, the antiderivative is $k^{-1}e^{kx}$ and the equation to solve is $e^{2k}-1=k$, subject to $k\ne0$. This can be solved in terms of the Lambert W function: since $-2(k+1)e^{-2(k+1)}=-2e^{-2}$, $k=-\frac12W_n(-2e^{-2})-1$, with $W_n$ a branch of the Lambert W. Just don't choose the branch mapping $-2e^{-2}$ to $-2$, as that would give $k=0$. You don't say whether you want complex roots, but I guess you want $f$ to be a PDF. The only real root is $k\approx-0.796812$.