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.
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 $C=10^b$.