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Consider an exponential relationship of the form $y = KC^x$ where $K$ and $C$ are constants.

Express the exponential function $y = KC^x$ as a linear function and describe how you would obtain the coefficients $K$ and $C$ from the linear equation of best fit. Note that you do not need to enter anything onto Excel, just describe mathematically.

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  • $\begingroup$ Hint: $\ln y=\ln K+(\ln C)x$. $\endgroup$ Commented May 23, 2016 at 7:31
  • $\begingroup$ 1. $y = KC^x$ 2. $ln y = x ln KC$ 3. $ln y = x ln K + ln C$ 4. $(ln y - ln C) / x = ln K$ (1 to 4 are the steps so if i've made a mistake its easier to refer to) Is this correct so far? My next move would be to isolate K so K= .... Not sure what to do next.. Perhaps.. 5. $K = e^((ln y - ln C) / x) Yes, no?? $\endgroup$
    – Zantheor
    Commented May 23, 2016 at 7:47
  • $\begingroup$ I will write a comment or answer in a few hours. Apologies about the delay. $\endgroup$ Commented May 23, 2016 at 11:22

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We assume that $y$ is always positive. Let $z$ be the logarithm of $y$. Which base? It does not matter. I will use the natural logarithm $\ln$, but others might use the logarithm to the base $10$.

Note that if $y=KC^x$ then $z=\ln y=\ln K+(\ln C)x$.

Suppose that we are given values $x_1,x_2,\dots, x_n$ of $x$, and the associated values $y_1,y_2,\dots,y_n$ of $y$. These might be two columns in a spreadsheet. Calculate the values $z_1=\ln(y_1), z_2=\ln(y_2), \dots, z_n=\ln(y_n)$.

If the $y_i$ are (approximately) related to the $x_i$ by $y_i=KC^{x_i}$, then we have, approximately, $$z_i=\ln K+(\ln C)x_i.$$ This is a linear relationship. Use the usual method to find the line of best fit $z=a+bx$ to the data $(x_i,z_i)$, that is, $(x_i,\ln(y_i))$.

Then our estimate for $\ln K$ will be given by $a$, meaning that our estimate for $K$ is $K=e^a$. Similarly, our estimate for $C$ is given by $C=e^b$.

If instead we use $z_i=\log_{10}(x_i)$, the analyis is much that same, except that at the end we use $K=10^a$ and $C=10^b$.

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  • $\begingroup$ $K = e^a = e^((-ln C)xi)$ ??? $\endgroup$
    – Zantheor
    Commented May 24, 2016 at 4:51
  • $\begingroup$ @Zantheor: The first equation is right. The second is not. We want $\ln C=b$ so $C=e^b$. No $x_i$. $\endgroup$ Commented May 24, 2016 at 4:52
  • $\begingroup$ That didn't quiet format correctly for me.. But a = (- ln C)xi..?? $\endgroup$
    – Zantheor
    Commented May 24, 2016 at 4:52
  • $\begingroup$ How do I write that though without the b variation, with the lettering etc used from the start of the post? I need to be able to show my working out $\endgroup$
    – Zantheor
    Commented May 24, 2016 at 5:42
  • $\begingroup$ I showed the working out in detail, apart from the finding of $a$ and $b$, which is obtained by finding the line of best fit to the set of points $(x_i,\ln(x_i))$. Then I showed how, once we have found $a$ and $b$, we find $K$ and $C$. That's what the question asked for. $\endgroup$ Commented May 24, 2016 at 5:48

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