# Find expotential function from two points

Clever people on this place, I'm having trouble with this, and I'm not able to see why what I'm doing is wrong... Here are two points:

$(3,1)$, $(-1,16)$

And this is what my calculations are:

First, I'll find $a$:

$a = (x_2-x_1)\sqrt{\frac{y2}{y1}}= (-1 + (-3) )\sqrt{\frac{16}{1}}\\\Leftrightarrow \\a = -4\sqrt{\frac{16}{1}}n \\\Leftrightarrow \\a = -4\sqrt{\frac{16}{1}} \Rightarrow a=−16$

Then, I can find $b$:

$b = (\frac{y1}{a^x1}))= \frac{1}{-16^3}\\ \Rightarrow b=−0.000244$

This is wrong, my book says that the answer is $f(x) = 8(2^{-x})$

What's wrong, and what should be changed here? I'm not the biggest math professor, but i hope you can help me aswell.

What i need to know is how the final function can be, as it says in the book - in this case, it's $f(x) = a (b^x)$

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You need to be a bit clearer about what you are trying to do. Are you trying to find $a$ and $b$ such that the function $f(x)=a\times b^x$ passes through your two points? –  Matthew Pressland Nov 20 '12 at 17:06
yes, exactly .. –  Frederik Witte Nov 20 '12 at 17:07
OK, great. Ideally you should edit your question to include this information. –  Matthew Pressland Nov 20 '12 at 17:08
I will, thanks for telling –  Frederik Witte Nov 20 '12 at 17:08
I assume the function you're trying to fit is $y=ba^x$. Where did you get $a=(x_2-x_1)/\sqrt{y_2/y_1}$ from? –  Rahul Nov 20 '12 at 17:10

You know that $f(3)=1$ and $f(-1)=16$, so as $f(x)=a\times b^x$, you have:

\begin{align*} ab^3&=1\\ ab^{-1}&=16 \end{align*}

Now we can cancel the $a$s by dividing:

$$b^4=\frac{ab^3}{ab^{-1}}=\frac{1}{16}$$

So one choice of $b$ is $b=\frac{1}{2}$, and then you can check that to satisfy the two equations you must take $a=8$. However, taking $b=-\frac{1}{2}$ and $a=-8$ also works, and there are two more choices where $a$ and $b$ are complex numbers.

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you sir, you are a genius - let your math soul bring you succes in the futute! –  Frederik Witte Nov 20 '12 at 17:19
I believe the generic solution is to take the base b logarithm of the equations that yields a linear regression problem, a and b can be found from the solution. en.wikipedia.org/wiki/Nonlinear_regression –  peterm Nov 20 '12 at 17:24