If you write as $\lambda = b - \mu$ then you can rewrite the exponential distribution as $f(x;b,\mu_ = (b-\mu)e^{-(b-\mu)x}$.

Now, you have access to iid sample $x_{1}, x_{2},..., x_{n},$ you can write the likelihood function

$$L(\mu|b,x_{1}, x_{2},..., x_{n}) = \prod_{i=1}^{n}(b-\mu)e^{-(b-\mu)x_{i}}$$

then $l(\mu|b,x_{1}, x_{2},..., x_{n}) = log(b-\mu)^{n} - (b-\mu)\sum_{i=1}^{n}x_{i}$, for which we can take the first derivative and equate it to zero so we can maximize it with respect to $\mu$.

$$\frac{dl(\mu|b,x_{1}, x_{2},..., x_{n})}{d\mu} = -\frac{n}{b-\mu}+\sum_{i=1}^{n} x_{i} = 0 \Rightarrow \mu = b - \frac{n}{\sum_{i=1}^{n}x_{i}}, \ \ \mu < b, \ \ b > 0 $$

If you write as $\lambda = b - \mu$ then you can rewrite the exponential distribution as $f(x;b,\mu_ = (b-\mu)e^{-(b-\mu)x}$.

Now, you have access to iid sample $x_{1}, x_{2},..., x_{n},$ you can write the likelihood function

$$L(\mu|b,x_{1}, x_{2},..., x_{n}) = \prod_{i=1}^{n}(b-\mu)e^{-(b-\mu)x_{i}}$$

then $l(\mu|b,x_{1}, x_{2},..., x_{n}) = log(b-\mu)^{n} - (b-\mu)\sum_{i=1}^{n}x_{i}$, for which we can take the first derivative and equate it to zero so we can maximize it with respect to $\mu$.

$$\frac{dl(\mu|b,x_{1}, x_{2},..., x_{n})}{d\mu} = -\frac{n}{b-\mu}+\sum_{i=1}^{n} x_{i} = 0 \Rightarrow \mu = b - \frac{n}{\sum_{i=1}^{n}x_{i}} = b - \frac{1}{\bar{x}}, \ \ \mu < b, \ \ b > 0 $$