Remember, that our exponential curve is in this form,
Let's take the natural log of both sides of the equation.
Recall the properties of logarithms that state
Ln[a b] = Ln[a] + Ln[b], and ,
Ln[e^x] = x.
Therefore, from equation (1) we get,
Ln[y] = Ln[a] + Ln[e^(bx)]
Ln[y] = Ln[a] + b x
Ln[y] = b x + Ln[a].
This looks like the equation of a line,
Ln[y] = b x + Ln[a]
(y) = (m x) + (b).
Assuming that the best fitting line goes through the point (xmean, ymean) for a line, we can show that it is the same point for an exponential except we use the mean of Ln[y] instead of y.
The mean data point of this equation would be the mean of x and the mean of Ln[y]. However, this is for a line and not for the curve. We are interested in this value in terms of y. To find this we need to raise e to the mean of Ln[y] to get back to the curve. Therefore, the point that must be on the best fitting curve of an exponential function is the point, (xmean, e ^ (mean of Ln[y])).