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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? –  Matt 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. –  Matt 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
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1 Answer 1

up vote 1 down vote accepted

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
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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. en.wikipedia.org/wiki/Nonlinear_regression –  peterm Nov 20 '12 at 17:24
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