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!
On
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.
On
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 $(0,2)$.
On
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.
On
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: $$\begin{align} x&=W\Big(-\frac{1.9}{2e}\Big)\qquad{\mathrm{recall \ \ that\ -2x=2+a}}\\ 2+a&=-2W\Big(-\frac{1.9}{2e}\Big)\\ \end{align}$$
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:
you can write $$-e^{-a/2}\left(1+\frac{1}{2}a\right)=-\frac{95}{100}$$ and then a numerical method will help you!