Help solving system of exponential equations I'm trying to design and 3D print a nozzle for gas pressure measurements. I'd like the inside of the nozzle to follow an exponential curve (probably not necessary, but I'd like to attempt this), but I can't solve the equations to set it up. I'm not asking for the final answer, just some help getting unstuck.
I'm designing the nozzle on the XZ plane (x: horizontal, z:vertical), in the upper left-hand quadrant (x negative, z positive). The exponential line has to pass through the (x,z) points (-9.02,0) and (-1.5,21) to fit the rest of the nozzle. When I've got the exponential part done, I can revolve the sketch in the XZ plane around the Z axis to make the 3D model.
So far, I've set up a pair of exponential equations, using $z=ab^x$:
Point 1: $0=ab^{-9.02}$
Point 2: $21=ab^{-1.5}$
This is as far as I can take it. For Point 1, it seems like I should be able to say $a=0$ because any base raised to any power times zero is zero, but I'm not sure of this.
I'm grateful for anyone who can suggest how to proceed from this point. Do I even have the two equations set up correctly?
Thanks so much for any help or advice on my problem.
-Kevin
Edit 1:
I was able to implement your equation in Octave and view the graphs. It works great; meeting the endpoint requirements, and changing shape based on the value of $b$. However, after all this, the shape is convex, where I wanted concave. I'll have to work on this some more next week to see how I can transform it.
Two graphs: Left is function, right is shape desired
Thanks, again, for all your help.
-Kevin
 A: The problem is that $0=ab^{-9.02}$ does not give a nice exponential curve; instead you end up with the constant 0 function.
You might have more luck with an offset exponential function, i.e. $z=ab^x+c$. Now we can solve the problem, leaving one unknown $b$:

*

*Point 1: $c = -ab^{-9.02}$

*Point 2: $c = 21 - ab^{-1.5}$
Thus we get $a=\frac{21}{b^{-1.5}-b^{-9.02}}$.
From Point 1 we then get $$c=-ab^{-9.02} = -\frac{21}{b^{-1.5}-b^{-9.02}} b^{-9.02} =\frac{21}{1-\frac{b^{-1.5}}{b^{-9.02}}}$$
Filling this in in $z=ab^x+c$ then gives the expression $$z=\frac{21b^x}{b^{-1.5}-b^{-9.02}}+\frac{21}{1-\frac{b^{-1.5}}{b^{-9.02}}}$$
You can then choose the parameter $b$ to be any number that is larger than 0 and not equal to 1  and look at the resulting curve to see if you like it. For example with geogebra
A: @student91, thanks, again.
I follow where you got your equation for $a$. Since $c=-ab^{-9.02}$ and $c=21-ab^{-1.5}$ therefore:
$$-ab^{-9.02} = 21-ab^{-1.5}$$
I can then solve for $a$ and get
$$a=21/(b^{-1.5}-b^{-9.02})$$
However, I can't see how to get your second equation, $$c=\frac{21}{1-\frac{b^{-1.5}}{b^{-9.02}}}$$
Could you elaborate on that, please?
Also, I don't understand your statement "You can then adjust the parameter $b$ to suit your needs." Are you saying that $b$ can take on any value? For instance, if I set $b=1$, then 1 to any power = 1 (correct?). The equation then becomes:
$$
\begin{align}
z&=\frac{21}{0} + \frac{21}{1-\frac{1}{1}} \\
&=\frac{21}{0} + \frac{21}{0}
\end{align}
$$
I sure wish I had done better in math in school.
Thanks, again, for all your help and advice.
-Kevin
