Solving equations with e on the form $1.22e^{0.015x} - 0.22 e^{-6x} = 3$ I'm wondering how equations of the form like: 
$1.22e^{0.015x} - 0.22  e^{-6x} = 3$
can be solved for $x$ without the use of an advanced calculator or plotting tool?
I tried substituting $e^x$ by u and solving for u, however the resulting expression is also complicated with several complex answers, which leads me to guess that this might not be the smartest approach.
Any ideas if there are simpler ways or are there no "shortcuts" for these types of equations?
 A: Your equation is not going to have an analytical solution. However, as other answers suggested you can neglect one term to have an equation which has now an analytical solution. But of course you need to check that the solution of this new equation is close to the solution of the original equation. Note that this original solution is unique by considering the derivatives and the limits at $\pm \infty $.
So if you neglect the term $ -0.22 e^{-6x}$ the equation becomes $ 1.22e^{0.015x} = 3$ which gives you as an approximate solution
 $ x_0 = \frac{\log \left( \frac{3}{1.22} \right) }{0.015} \approx 59.984 \dots $
Now let's set $ x = x_0 + \epsilon $ and look how to evaluate the error $\epsilon $ assuming $ \epsilon \ll x_0 $.
Let's $a=1.22$, $b=0.015$, $c=-0.22$, $d=-6$ and $f = -3 $ so you equation is :
$$a\cdot e^{bx} +c\cdot e^{dx} +f = 0$$
And we know :
$$ a\cdot e^{bx_0} + f = 0$$
Therefore :
$$a\cdot e^{b(x_0 + \epsilon)} +c\cdot e^{d(x_0 + \epsilon)} +f = 0$$
$$a\cdot e^{bx_0} e^{b \epsilon} +c\cdot e^{dx_0}e^{d\epsilon} +f = 0$$
$$c\cdot e^{dx_0}e^{d\epsilon} = f \cdot \left(e^{b \epsilon} -1 \right)$$
Only considering the first order in $ \epsilon$ we have :
$$e^{d\epsilon} \approx 1 \text{ and }e^{b \epsilon} - 1 \approx  \frac{(b\epsilon)^2}{2}$$
So in the end :
$$ \epsilon \approx \frac{1}{b} \sqrt{ \frac{2c \cdot e^{dx_0}}{f}} \approx 1.7979998 \cdot 10^{-77} $$
So even if we neglected one term in the original equation, we know that the our approximate result is quite close the true solution.
A: let $$f(x)=\frac{61}{50}~e^{\frac{3}{20}x}-\frac{11}{50}e^{-6x}-3$$ then $$f'(x)={\frac {183}{1000}{{\rm e}^{{\frac {3\,x}{20}}}}}+{\frac {33\,{{\rm e}
^{-6\,x}}}{25}}>0$$ and we have $$\lim_{x\to \infty}f(x)=+\infty$$ and $$\lim_{x\to -\infty}f(x)=-\infty$$ and we get that our equation has a unit determed solution. 
A: First note that 6/.015= 400 so if we let y= .015x, this equation becomes $1.22e^y- 0.22e^{-400y}= $1.22e^y- 0.22(e^y)^{-400}= 3.  
Now, let z= e^y and we have $1.22z- 0.22z^{-400}= 3$ which is equivalent to the polynomial equation $1.22z^{401}-3z^{400}- 0.22= 0$.
