Estimating the parameter $\beta$ The lifetime of computer monitors has a exponential distribution where the expected value can be written as:
$\mu(s) = \frac{\beta}{s}$
Where $s$ is how bright the monitor is and both $s$ and $\beta$ are >0
How do I estimate the maximum-likelihood of parameter $\beta$?
I know that the log-likelihood for a exponential distribution is:
$l(\lambda) = n(\ln(\lambda) - \lambda\bar{t})$
And taking the derivative and equals it to zero yields:
$\frac{dl(\lambda)}{d\lambda} = n(\frac{1}{\lambda} - \bar{t}) = 0$
$\bar{t} = \frac{1}{\lambda}$
Here is where I get stuck.
EDIT 1
I have data for both the lifetime of the computer monitors and their brightness.
 A: To estimate the $\beta$ by the maximum likelihood method, let $Y_1,\ldots, Y_n$ be the sample of lifetimes, with $Y_i\sim \text{Exp}(\beta/s_i)$, s.t. $E(Y_i) = \beta/s_i$ and independently for each $i$, where $s_i$ is the monitor brightness.
Then the likelihood function for $\beta$ is given by
$$
L(\beta) = \prod_{i=1}^n (s_i/\beta) e^{-\frac{1}{\beta}s_i Y_i} \propto \beta^{-n} e^{-\frac{1}{\beta}\sum_is_iY_i}.
$$
I leave the rest to you...
A: Assise from using a direct derivation, leading to a closed form expression, you can do this with a generalized linear model a description is given in the question
Fitting exponential (regression) model by MLE?
With that approach you have more flexibility (e.g. change the function $\mu(s)$)
Example
In R you would use the function glm with the family gamma and specify a dispersion parameter dispersion=1 (the exponential distribution is a special case of the gamma distribution for which the dispersion parameter is equal to 1).
### generate and plot data 

set.seed(1)
n = 100
x = runif(n,0,10)
mu = 2/x
rate = 1/mu
y = rexp(n,rate)
plot(x,y)

#### fit glm model with Gamma distribution that has k=1 (ie. the exponential distribution)

fit = glm(y ~ 0+x, family = Gamma())
summary(fit,dispersion=1)

# This gives as result 
# a rate parameter of lambda = 0.49868 x
# which relates to
# a mean that is mu = 1/lambda = 2.005294/x

A: $
L(\beta) = P(\mathrm{Data} \mid \beta) = \prod_i f(x_i)
$ and $\lambda = \frac{1}{\mu} = \frac{s}{\beta}$
Log is increasing so maximizing log is same as maximizing likelihood:
$\ell(\beta) = \sum \log f(x_i).$
$
\log f(x_i) = \log \lambda e^{-\lambda x_i} = \log s - \log \beta  - \frac{s}{\beta}x_i$
and

 \frac{d}{d \beta} \log f = -\frac{1}{\beta} - \frac{s}{\beta^2}x_i.

So,

 \frac{d}{d \beta}\ell(\beta) = -\frac{n}{\beta} - \frac{s}{\beta^2}\sum x_i = 0

gives

 \hat{\beta} = \frac{s}{n}\sum x_i = s \bar{x}.

Note: the second derivative of $\ell$ is positive since $x_i, s, \beta>0$ so  LL is concave.
