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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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    I'm sorry for being confusing - 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) =$a$*$b$^ x2012-11-20

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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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    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. http://en.wikipedia.org/wiki/Nonlinear_regression2012-11-20