Three points --> how to build function?

Are you sure it's a logarithmic problem?

This sounds like a regression problem. The first step in a regression problem is to define the fitting equation. When you say this should be logarithmic, are you specifically thinking of something like y=a*ln(x) + b or some other logarithmic function? Whatever your function is, are you expecting to use a standard least squares regression or a regression based on some other objective function?

If you want a standard least squares regression and your desired function is "linear" (such as y=mx+b or y=mln(x)+b or a polynomial or any other linear or linearizable function), then you can use the LINEST() function to perform the regression: https://support.office.com/en-us/art...rs=en-US&ad=US Enter your y's and x's (and any transformations needed), then enter the LINEST() function to obtain the regression coefficients.

If you want some other objective function or your desired function is non-linear, then you can use Excel's built in Solver to perform the regression. Enter your data, enter some reasonable guess for the parameters of the function, compute your desired objective function (the function that will be minimized to obtain the regression such as sum((y-yequation)^2)), then call Solver and tell it to set target cell (objective function) to a minimum by changing your parameters.

I realize that is a quick overview. Let us know in more detail what you are trying to do and exactly what steps you need help with and we'll help as best we can.

Do you possibly have more points you could supply? Currently, your data set does not seem to follow a normal logarithmic trend, rather a linear (or even power series) trend. We can estimate your data for a y-value of 385, but I am assuming it would have a rather large error associated with it (mainly due to the small sample size of the regression).

So using FORECAST for a linear regression, I received a y-value of 6851.831 for an x-value of 365. Using this for your plot, you can obtain a logarithmic equation of y=1146.7ln(x)-1313.2

This has an R²=0.7452 (in case you didn't know, the closer to 1.0000 the value is, the more accurate the trend fit).

You can take a 75% accuracy for what it's worth. Bottom line is if more data is provided, a more precise trend can be created.

PS: The FORECAST() data is pretty close to the trendline data created by hand. The results were 98.7% close, so hopefully that approach is adequate. I attached a sample workbook with what I did to show you my thought process. A useful guide to creating your own forecasts can also be seen here http://spreadsheetpage.com/index.php...line_formulas/

Hope this helps!