TITLE: Solving exponential equation by hand?
QUESTION [7 upvotes]: I was wondering how to solve the following exponential equation for a by hand:
$$-e^{-a/2}-\frac{1}{2}ae^{-a/2} = -0.95$$
I'm able to find a solution with computer software, but am stuck when it comes to solving it algebraically by hand. Any help would be appreciated!

REPLY [10 votes]: We can use the Lambert $W$ function to solve the equation "by hand".
Notice that your equation can be written as:
$$2+a-1.9e^{\frac a2}=0 \Rightarrow \color{red}{1.9e^{\frac a2}=2+a}$$
Now we let $-2x=2+a$ and $a=-2x-2$ and thus we have:
$$-2x=1.9\exp\Big(\frac{-2x-2}{2}\Big)$$
$$-2x=1.9\exp(-x-1)\Rightarrow\color{red}{-2x=\frac{1.9}{e^x\cdot e}}$$
The last equation can be rewritten as:
$$x\cdot \color{red}{e^x}=\frac{1.9}{\color{red}{-2}\cdot e}$$
And now we have the function $xe^x=k$, which can be written as $x=W(k)$, and thus we have:
$$\begin{align}
x&=W\Big(-\frac{1.9}{2e}\Big)\qquad{\mathrm{recall \ \ that\ -2x=2+a}}\\
2+a&=-2W\Big(-\frac{1.9}{2e}\Big)\\
\end{align}$$
$$\therefore\bbox[8px, border:2px solid black]{a=-2-2W\Big(-\frac{1.9}{2e}\Big)}$$
Now, it is debatable whether the Lambert-$W$ function constitutes an explicit solution. However, we can solve this numerically thru NR, using the recurrence relation: 
$${a_{n+1}=-2+a_n-\frac{a_n}{1-0.95e^{\frac a2}}}$$
For approximately close $a_0$ and after sufficient iterations, the solution over $\mathbb{R}$ would be:
$$\bbox[8px, border: 2px solid black]{a=\begin{cases}
-0.574097\\
0.710723\\
\end{cases}}$$