I have charted 2 sets of 11 data points each on a graph and want to create an equation that defines each graph so I can input any single axis data and it will find the corresponding axis point of intersection
I have charted 2 sets of 11 data points each on a graph and want to create an equation that defines each graph so I can input any single axis data and it will find the corresponding axis point of intersection
Not much to go on. Other than using the chart to visualize the process, I would probably not do any of this in the chart. If I understand what you are trying to do:
1) Choose the desired equation form. I am not certain exactly what you want for a "logarithmic equation", but I assume you know what you are looking for. As an example, I will assume you are looking for y=m*ln(x)+b
2) Use the LINEST() function to perform the desired regression. For the example, =LINEST(known_ys,LN(known_xs)) https://support.office.com/en-us/art...a-fa7abf772b6d
3a) Compute the desired y at a given value for x.
3b) Use algebra to invert the function, then compute the desired x at the given value for y, or use Solver/Goal Seek/other numerical algorithm to find x at a given y.
Originally Posted by shg
Does a logarithmic trendline give a sufficiently accurate fit for what you need?
Entia non sunt multiplicanda sine necessitate
i dont know. the graph is pretty smooth.
i think what I am trying to do is reverse engineer the graph so I can use a formula to predict any y point on graph given an x point.
I can supply the data if you like
no i do not
Right-click the data series on the chart, select Add Trendline, and follow your nose from there.
I don't know if using a trendline is going to work. It is a smooth concave shaped curve and the trendline is straight. I want to predict points on the curve not the trendline
What trendline type (linear, polynomial, logarithmic, power, etc.) did you choose? It sounds like you chose "linear", when you should have chosen something else.
Post a workbook, Brian?
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