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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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    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?2012-11-20
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    yes, exactly ..2012-11-20
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    OK, great. Ideally you should edit your question to include this information.2012-11-20
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    I will, thanks for telling2012-11-20
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    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?2012-11-20
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    i googled it, and found it on another math site2012-11-20
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    I see your edit, but it isn't enough. You've told us you want to find "the function", but you haven't said what the form of the function actually is. We're all guessing.2012-11-20
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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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    you sir, you are a genius - let your math soul bring you succes in the futute!2012-11-20
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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