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I really need help solving this problem, and I hope I came to the right place (this is my first time on this site)! For a certain exponential function, I was given the points (-10, 1.9968), (-9,1.984), (-8, 1.92), (-7,1.6) and (-6,0). By the points, I was able to make some conclusions for the exponential function:

y = -a(5)^[k(x-d)] +2,

the b-value is 5, the c-value (or the asymptote) is 2, and the a-value is negative, as the function portrays to be decaying. I am in need of finding the a, k and d values of this exponential function, as I need it for a math assignment due soon. I would like to know if this is high school level math, as I may be interpreting the question differently. Thank you very much and I appreciate all your replies.

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    $\begingroup$ least squares between log y and x $\endgroup$ – Narasimham Jan 19 '18 at 4:37
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You have an extra parameter in your equation because $5^{-kd}$ can be absorbed into $a$, so now your equation is $y=-a5^{kx}+2$. If you subtract $2$ and take the log you get $\log_5(2-y)=\log_5(a)+kx$. This is a straight line in $x$ vs $\log_5(2-y)$ and you should be able to find $k$. I plotted the points and they look to fit a straight line nicely.

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  • $\begingroup$ Hi there, thanks for your efforts in helping me! I was able to algebraically discover that the k value is 1. How exactly would I find the remaining two variables; a and d variables. $\endgroup$ – ProgrammerNoob Jan 20 '18 at 2:47
  • $\begingroup$ Then $a$ and $d$ are equivalent just as I said. You can just set $d$ to zero and find the proper $a$. $d$ just makes a factor of $5^{-kd}$, and given $k=1$ this is $5^{-d}$. It is a multiplicative factor just like $a$, so the two of them are only one variable. If there is no noise in your data, you can just take one point and evaluate $a$. Otherwise you can compute $a$ from each point and take some sort of average. $\endgroup$ – Ross Millikan Jan 20 '18 at 3:41

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