I'm trying to solve for $\mu_k$ the following equation:
$$\ln\left(\frac{\mu_k}{1-\sum_{j}\mu_j}\right) = \eta_k$$
and the book Pattern Recognition and Machine Learning says that to solve for $\mu_k$ first sum both sides over $k$ and then rearrange and back-substitute to give:
$$\mu_k = \frac{e^{\eta_k}}{1 + \sum_{j} e^{\eta_j}}$$
I'm not quite sure what he means, but think he's referring to my work below. If not, what is the correct way to solve this?
$$\begin{align} \sum_k \ln\left(\frac{\mu_k}{1-\sum_j \mu_j}\right) = & \sum_k \eta_k & \text{(sum both sides over $k$)} \\ \sum_k \frac{\mu_k}{1-\sum_j \mu_j} = & \sum_k e^{\eta_k} & \text{(exponentiate both sides)} \\ \frac{1}{1-\sum_j \mu_j} \sum_k \mu_k = & \sum_k e^{\eta_k} \\ \sum_k \mu_k = & \left(1-\sum_j \mu_j\right) \sum_k e^{\eta_k} \\ \sum_k \mu_k = & \sum_k e^{\eta_k} - \sum_j \mu_j \sum_k e^{\eta_k} \\ \sum_j \mu_j \sum_k e^{\eta_k} + \sum_k \mu_k = & \sum_k e^{\eta_k} \\ \sum_k \mu_k \left( \sum_j e^{\eta_j} + 1\right) = & \sum_k e^{\eta_k} & \text{(factor and swap indices... valid?)}\\ \sum_k \mu_k = & \frac{\sum_k e^{\eta_k}}{\sum_j e^{\eta_j} + 1} \\ \mu_k = & \frac{e^{\eta_k}}{1 + \sum_j e^{\eta_j}} & \text{(remove sum over $k$)} \\ \end{align}$$
UPDATE: The proper solution (last step above is invalid) to this problem (thanks Clement C.) would be:
$$\begin{align} \frac{\mu_k}{1-\sum_j \mu_j} = & e^{\eta_k} & \text{(exponentiate both sides)} \\ \mu_k = & \left(1-\sum_j \mu_j\right) e^{\eta_k} & \text{Note: $1-\sum_j \mu_j = \frac{\mu_k}{e^{\eta_k}}$} \\ \sum_k \mu_k = & \left(1-\sum_j \mu_j\right) \sum_k e^{\eta_k} & \text{(sum both sides over $k$)} \\ \sum_k \mu_k = & \sum_k e^{\eta_k} - \sum_j \mu_j \sum_k e^{\eta_k} \\ \sum_j \mu_j \sum_k e^{\eta_k} + \sum_k \mu_k = & \sum_k e^{\eta_k} \\ \sum_k \mu_k \left( \sum_j e^{\eta_j} + 1\right) = & \sum_k e^{\eta_k} & \text{(factor and swap indices)}\\ \sum_k \mu_k = & \frac{\sum_k e^{\eta_k}}{\sum_j e^{\eta_j} + 1} \\ 1 - \sum_k \mu_k = & 1 - \frac{\sum_k e^{\eta_k}}{\sum_j e^{\eta_j} + 1} & \text{(subtract LHS and RHS from 1)} \\ 1 - \sum_k \mu_k = & \frac{1+\sum_j e^{\eta_j}}{1+\sum_j e^{\eta_j}} - \frac{\sum_k e^{\eta_k}}{1 + \sum_j e^{\eta_j}} \\ 1 - \sum_k \mu_k = & \frac{1}{1 + \sum_j e^{\eta_j}} \\ \frac{\mu_k}{e^{\eta_k}} = & \frac{1}{1 + \sum_j e^{\eta_j}} & \text{substitute: $1-\sum_j \mu_j = \frac{\mu_k}{e^{\eta_k}}$} \\ \mu_k = & \frac{e^{\eta_k}}{1 + \sum_j e^{\eta_j}} \end{align}$$
Yes, $$\sum_k \mu_k \left( \sum_j e^{\eta_j} + 1\right) = \sum_k e^{\eta_k} \quad \Longleftrightarrow \quad \sum_k \mu_k = \frac{\sum_k e^{\eta_k}}{\sum_j e^{\eta_j} + 1}, $$ since $\sum_j e^{\eta_j} + 1$ does not depend on $k$.
You are actuall using some rule like $$s=\sum_k ca_k \quad \Longleftrightarrow \quad \frac{s}{c}=\sum_k a_k . $$
And with the same arguments we have $$\sum_j \mu_j \sum_k e^{\eta_k} + \sum_k \mu_k = \sum_k e^{\eta_k} \quad \Longleftrightarrow \quad \sum_k \mu_k \left( \sum_j e^{\eta_j} + 1\right) = \sum_k e^{\eta_k}$$ because $$\sum_k \mu_k =1\cdot \sum_j \mu_j$$ and we use the well known rule $$\sum_j a\mu_j + \sum_j b \mu_j = \sum_j (a+b)\mu_j.$$ Here $a=\sum_k e^{\eta_k}$ and $b=1$.