Conditional probability with MLE of Poisson variable I'm having some trouble with this study question and would appreciate any help. This may be a duplicate but I have not been able to find any others.

Question: Leaves of plants are examined for bugs. The number of bugs on a leaf follows a Poisson distribution with mean $\mu$. However, many leaves have no bugs because they are unsuitable for feeding and not simply because of the chance variation allowed by the Poisson law. Therefore the empty leaves are not counted.
a) Find the conditional probability that a leaf contains $x$ bugs, given that it contains at least one.
b) Let $x_i$ be the number of bugs found on leaf $i$ (after the leaves with zero bugs are ignored).
Show that the MLE of $\mu$ satisfies the equation
$$\hat{\mu}=\bar{x}(1-e^{-\hat{\mu}}).$$
c) Determine $\hat{\mu}$ numerically for the case $\bar{x}=3.2.$

My attempt:
a) If X is the number of bugs on a leaf and Y is the number of observed bugs on a leaf, then the probability is given by 
\begin{align}
P(Y=x)&=P(X=x|X>0) \\
&=\frac{P(X=x)}{P(X>0)} \\
&=\frac{P(X=x)}{P(X>0)} \\
&=\frac{P(X=x)}{1-P(X=0)} \\
&=\frac{P(X=x)}{1-e^{-\mu}} \\
\end{align}
b) Having some trouble with this. First I found the regular MLE of $\hat{\mu}$ for Poisson and got $\bar{x}=\hat{\mu}$, so I'm assuming I need to use $$\frac{\mu}{1-e^{-\mu}}$$ in some way? Is this what my answer to a) is supposed to be? I tried finding the MLE of $\hat{\mu}$ again using this instead of $\mu$ but couldn't get the answer I needed.
c) Not sure what to do here... 
 A: Assuming $X\sim\mathsf{Poisson}(\mu)$, the probability that a leaf contains $x$ bugs given that it contains at least one is
\begin{align}
P(X=x\mid X\ge 1)&=\frac{P(X=x)}{P(X\ge 1)}\mathbf1_{x\in\{1,2,\ldots\}}
\\&=\frac{e^{-\mu}\mu^x}{x!(1-e^{-\mu})}\mathbf1_{x\in\{1,2,\ldots\}}
\end{align}
For finding MLE of $\mu$, this conditional pmf should be used for your likelihood. You cannot say that the MLE of $\mu$ is simply $\bar x$ because here you are considering a Poisson distribution which is truncated at zero.
For $\mu>0$, the likelihood function given the sample $x_1,\ldots,x_n$ is
\begin{align}
L(\mu)&=\prod_{i=1}^n \frac{e^{-\mu}\mu^{x_i}}{x_i!(1-e^{-\mu})}\mathbf1_{x_i\in\{1,2,\ldots\}}
\\&=\left(\frac{e^{-\mu}}{1-e^{-\mu}}\right)^n \frac{\mu^{n\bar x}}{\prod x_i!}\mathbf1_{x_1,\ldots,x_n\in\{1,2,\ldots\}}
\end{align}
Log-likelihood is of the form
$$\ell(\mu)=n\ln\left(\frac{e^{-\mu}}{1-e^{-\mu}}\right)+n\bar x\ln\mu+h(x_1,\ldots,x_n)$$
Now taking the derivative of $\ell(\mu)$ and setting it to zero and so on, you get  $$\hat\mu=\bar x(1-e^{-\hat\mu})$$, where $\hat\mu$ is the desired MLE of $\mu$.
The last equation cannot be solved analytically for $\hat\mu$. So you are asked to use numerical methods (e.g., Newton-Raphson) to find $\hat\mu$ when $\bar x=3.2$.
