$$f(t;τ) = \frac{1}{\tau}\,e^{-t/\tau},$$ where $\tau$ is the mean time to failure.
$$L(\tau) = \prod_{i=1}^r \frac{1}{\tau} \,e^{-t_i/\tau} \prod_{j=1}^{n-r} e^{-t_r/\tau}.$$
Find the maximum likelihood estimator for $\tau$.
I'm having trouble taking the $\log$ of $L(\tau).$
On
Look at Wikipedia for the parameterization of the exponential distribution according to the mean (not rate.) For intuition, it might help to start by showing (integration by parts) that $\mu = E(T) = \tau.$ It might also help if you were to fully specify the density function as holding "for $t > 0.$" I agree with @MichaelHardy that your likelihood function is wrong.
[It is unclear to me whether you are supposed to use only one observation ($n=1$). But if so, why is there an $n$ in your proposed likelihood function, and what is $r?$. You say nothing about your data.]
It is shown there than the MLE for the rate $\lambda = 1/\tau$ is $\hat \lambda = 1/\bar T,$ where $\bar T = \frac 1 n \sum_{i=1}^n T_i.$ By invariance $\hat \tau = \hat \mu = \bar T.$ (If $n=1,$ then use $T$ instead of $\bar T.$)
The figure below shows two likelihood functions for the case $\tau = 4.$ In the upper panel $n=100$ observations have $\hat \tau = \bar T = 3.96.$ In the lower panel $n = 20$ observations have $\hat \tau = \bar T = 3.20.$ In each case, the MLE $\hat \tau$ occurs at the maximum of the likelihood function. In the first case the curvature of the likelihood function is greater near the maximum, making a 'sharper' approach to the MLE. This is related to the fact that estimates based on 100 observations tend to be better than ones based on 20 observations.
Note: In case you are familiar with R statistical software, here is the R code for making the figure:
m = 1000; tau = seq(.01, 10, length=m) # value of tau to plot
set.seed(228)
x1 = rexp(100, 1/4); mean(x1); x2 = rexp(20, 1/4); mean(x2)
## 3.96023 # MLE 1
## 3.199345 # MLE 2
par(mfrow=c(2,1)) # two panels per plot
like=numeric(m)
for(i in 1:m) {
like[i] = prod(dexp(x1, 1/tau[i])) }
plot(tau, like, lwd=2, col="blue", type="l", main="Likelihood Function for n=100")
abline(h=0, col="green3"); abline(v=mean(x1), col="red", lty="dashed")
like=numeric(m)
for(i in 1:m) {
like[i] = prod(dexp(x2, 1/tau[i])) }
plot(tau, like, lwd=2, col="blue", type="l", main="Likelihood Function for n=20")
abline(h=0, col="green3"); abline(v=mean(x2), col="red", lty="dashed")
par(mfrow=c(1,1))
The likelihood function you wrote would correspond to a sample $\boldsymbol t$ of size $n$ in which $0 \le r \le n$ observations are known event times, and the remaining $n-r$ observations are right-censored at time $t_r$, where event times are assumed to be independent exponential random variables with rate $1/\tau$. Then indeed we have $$\mathcal L(\tau \mid \boldsymbol t) = \prod_{i=1}^r \frac{1}{\tau} e^{-t_i/\tau} \prod_{j=1}^{n-r} e^{-t_r/\tau}.$$ Note that as you have written it, $t_r$ is both the censoring time and the $r^{\rm th}$ event time observed. If this is not your intent, you should use a different notation for the censoring time, say $t_c$. To this end, I will use this notation and you can extract your MLE as the special case when $c = r$.
Immediately, we may write $$\mathcal L(\tau \mid \boldsymbol t) = \tau^{-r} e^{-(\sum_{i=1}^r t_i)/\tau} e^{-(n-r) t_c/\tau},$$ ignoring indicator functions. The log-likelihood is then $$\ell(\tau \mid \boldsymbol t) = -r \log \tau - \frac{1}{\tau} \left(\sum_{i=1}^r t_i + (n-r) t_c \right).$$ The critical points of $\ell$ must satisfy $$0 = \frac{\partial \ell}{\partial \tau} = -\frac{r}{\tau} + \frac{K}{\tau^2},$$ where $$K = \sum_{i=1}^r t_i + (n-r)t_c,$$ hence $$\hat \tau = \frac{K}{r}.$$
As you should understand by now, the form of the likelihood depends on the structure of the data. Since you did not specify this structure in your question, the only way to proceed is to assume your expression for the likelihood is correct, and allow that to inform the data structure. But we have no way to know what your intended model was.