Solving exponential equation by hand? 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!
 A: As already said, numerical methods would be required.
However, consider that you look for the zero's of
$$f(a)=e^{-\frac a 2}\left(1+\frac{1}{2}a\right)-\frac{19}{20}$$ Using Taylor series built around $a=0$ you would get
$$f(a)=-\frac{1}{20}+\frac{a^2}{8}-\frac{a^3}{24}+\frac{a^4}{128}+O\left(a^5\right)$$ which can be solved using (ugly) radicals.
The solutions will be $\approx -0.574459$ and  $\approx 0.709325$ while the exact solutions would be $\approx -0.574097$ and  $\approx 0.710723$ which is not too bad.
Sooner or later, you will learn that, better than with Taylor series, functions can be approximated using Padé approximants. Using a $[2,2]$ one, we should have
$$f(a)=\frac{-\frac{1}{20}-\frac{1}{60}a+\frac{353 }{2880}a^2}{1+\frac{1}{3}a+\frac{7}{144}a^2}$$ and the numerator cancels for
$$a_ \pm=\frac{12}{353} \left(2\pm\sqrt{357}\right)$$ that is to say  $\approx -0.574315$ and  $\approx 0.710293$. We did not need to solve an ugly quartic polynomial to have something quite good.
A: you can write $$-e^{-a/2}\left(1+\frac{1}{2}a\right)=-\frac{95}{100}$$ and then a numerical method will help you!
A: You did the right thing to use  your calculator to solve  $$ -e^{-a/2}-\frac{1}{2}ae^{-a/2} = -0.95$$ 
Our current algebra is not sufficient to solve such equations by hand.
A: 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:
$$\bbox[8px, border: 2px solid black]{a=\begin{cases}
-0.574097\\
0.710723\\
\end{cases}}$$
A: If you want a solution ''by hand'' this can be only  an approximate solution and the approximation  depends on the work you are willing to do (and that a calculator can do in a much shorter time)  .
With a little work (and using your knowledge of elementary continuous functions):
Plot $y_1=1+\frac{1}{2}x$ and $y_2=0.95e^{\frac{x}{2}}$
You can easily see that for $x\ge 2$ or $x\le -2$ we have $y_2>y_1$ 
but, in a sub-interval  of $(-2 ,2)$ the value of $y_2$ becomes less than the value of $y_1$ because the two functions are continuous and $y_1(0)=1<0.95=y_2(0)$
So you can conclude that the equation $-e^{-a/2}-\frac{1}{2}ae^{-a/2} = -0.95$  has one solution in the interval $(-2,0)$ and another solution in $(0,2)$. 
