Specific value on Log Scale

[mombandre]

Hi guys. I'm new here.

I'm stucked in a very simple question:

How I'm supossed to find out the value 21,42? This value corresponds to 61,8% of the range from 9,17 to 34,94.

This is a logscale (and I actually dont know its basis).

 

[Dr.Peterson]

I know nothing about the context, but on a semilog graph, the labels are the values; the value of 21.42 is 21.42. You don't need to know the logarithm of that, as far as I can tell without knowing why you ask.

The percentages appear to refer to the logs, not the actual values. It's 61.8% of the way from bottom to top as actually measured on the graph.

Can you tell us more about what you want to do with the answer?

 

[JeffM]

This is a totally absurd presentation due to its lack of specifics.

If, however, that faint line is supposed to be a line of best fit, it is dubious because the residuals for small x are uniformly positive and the resuduals for large x are uniformly negative.

I had a friendly argument with SK about this recently. You can always find a line of best fit for any hypothesized type of functional relationship. It does not mean the the best fit is any good. Non-random residuals are an indicator that the line of best fit is actually a bad fit.

 

[Dr.Peterson]

Hi guys. I'm new here.

I'm stucked in a very simple question:

How I'm supossed to find out the value 21,42? This value corresponds to 61,8% of the range from 9,17 to 34,94.

This is a logscale (and I actually dont know its basis).

Looking back at this question, I realize I misinterpreted it. I apologize; your English is not as bad as I must have supposed it to be.

I think you were asking how to obtain the value 21.42 given the values 9.71 and 34.94, and the percentage 61.8%.

To generalize, suppose the lower and upper values are A and B. Then the actual positions on the graph are log(A) and log(B). Given a percentage P, we want that percentage of the range log(B)-log(A) = log(B/A), added to log(A): that is, the log of the desired position is log(A) + P log(B/A) = log(A*(B/A)^P). Therefore, the value at that percentage on the graph is the antilog of this, C = A*(B/A)^P.

In the example, A = 9.17, B = 34.94, and P = 0.618. Therefore, C = A*(B/A)^P = 9.17*(34.94/9.17)^0.618 = 9.17*(3.81)^0.618 =9.17*2.2858 = 20.96.

Possibly the numbers in the image are rounded slightly. Or else I've made a mistake somewhere.

But my question is, how did they choose the value 61.8%? Is it meant to be the golden ratio, or something else? And is it supposed to be significant?

 

[Cubist]

In the example, A = 9.17, B = 34.94, and P = 0.618. Therefore, C = A*(B/A)^P = 9.17*(34.94/9.17)^0.618 = 9.17*(3.81)^0.618 =9.17*2.2858 = 20.96.

Possibly the numbers in the image are rounded slightly.

The OP misquoted 9.17 instead of the 9.71 shown in the graph. Using the latter your calculation gives the expected result...

A = 9.71, B = 34.94, and P = 0.618. Therefore, C = A*(B/A)^P = 9.71*(34.94/9.71)^0.618 ≈.21.424