Log x axis help please

Hi,

I have a data
x values are 0,0.00125,0.0025,0.005,0.01,0.0625,0.125,0.25,0.5,1

and corresponding y values


100,36.04363183,39.50750344,38.44265851,33.42891946,33.48619236,45.09,41.32368779,41.4299366,40.73028774

If I want to chart them making my x axis in log scale in XY plot, the y value for x value of '0' doesnot seem to come up.

I understand the x log value should be more than 0.

How to make x values log in my case please. I would like to keep 0 as well as my first x value as it is control value for me.

Thanks

Thanks for the reply.
But the example is for only y axis.
I would like to convert x axis into log scale and my first value is x=0

So I don't know how to go.


Sincerely

Thanks.
Yes I have done it in XY graph.
See my first value on x axis is 0 followed by x=0.000625 and ..... the values given above.


When I convert X AXIS TO LOG scale I donot get the point (x,y)= (0,100)

I think there is no LOG 0?

Thanks

Log(0) is undefined, that's why it (0,100) can't be plotted. You will have to "defiine" log(0) for you application.

I'm not sure what you're trying to show with your plot, but you could use some arbitrary number smaller than the smallest x value. For example, instead of plotting (0,100), plot (1E-4,100). Then, if you apply a number format of 0.000 to the axis, the 1E-4 point will appear at 0.000. If you will, in this case you are simply "defining" log(0) to be log(a) where a is the smallest number that is meaningful to your application.

Tushar's flexible log axis tutorial applies to the x-axis as well, just replace references to the y-axis with x-axis. It's still the same principle. But, even using the flexible log axis, you still have to decide exactly what to do with the (0,100) point.

Mr Shorty,

Thank you very much. That is great and I used your method.
However I will try Tushar web later. You saved my enormous amount of time.



Thanks

Irr

Hi, I plotted my data as suggested my Mr Shorty. Great
However I am not able to fit a trend value.
The fact that i have converted my x values into log, does it affect the trendlines??

Please help me

Now you've introduced curve fitting into the problem. Excel can still calculate a trendline based on a log-x axis, so it isn't a problem with Excel.

Introducing curve fitting to the problem means you have to be very careful about changing a data point like I suggested. You can get very different trendlines depending on what you choose to substitute in for x=0.

The first step in any curve fitting problem is to determine what form you want for the function that will best represent the underlying real world process. Not knowing what the data represent, the only "trend" I see is y~40 (ignoring the (0,100) point).

What form of equation do you want? If you don't know the form, what properties do you want the equation to have?

That is a tough one. Off the top of my head, I don't see an obvious function that yields that kind of behavior. I'm sure you could get a high order polynomial to "fit" the data, but I would be real careful what such an equation would do in between the data points. The closest thing I came up with quickly was a decaying oscillation type equation (y=A^x*sin(x)) where A is less than 1.

Are you up to splitting the data set into 2 or 3 sections? Looking at the data on a linear axis, the data above 0.125 look like they could be fit by an exponential decay type model (A+B^-x) or an inverse function (A+Bx^-1). I don't know that these equation forms are readily represented by the forms available for chart trendlines. I would suggest you research (here and elsewhere) how to use the LINEST function to curve fit. You have much more flexibility in equation form (essentially anything that can be expressed in the form g(y)=a1*f1(x)+a2*f2(x)+a3*f3(x)...).

For ultimate flexibility, you can use Solver to get a best fit.

You could then try a cubic or quartic to fit the lower data. It also might be useful to "fix" the y intercept at 100.

Good luck and let us know how it goes and how we can help.

Thanks Mr shorty.

I will try and let you know anything comes out of it. Since I am not statitician I might need some time to fix this.

Thanks for your valuable time.

sincerely
Irr