I have a exponetial graph but how do I calculate results?

[Newbie_]

Homework Statement

I have a set of data by HPLC that will only fit in an exponential plot in excel.

If I was to use a linear graph I would use y=mx+c but the data will NOT fit into a linear plot.

Homework Equations

The R2 for the exponential graph = 0.9973

y=0.0044e(0.0017x)

where (0.0017x) is superscript.

The Attempt at a Solution

I have a value for my recovery that I can read off the graph but don't know how to convert this to ug (I can do this if the graph is linear).

I have plotted µg vs peak response.

Thanks!

 

[SammyS]

Homework Statement

I have a set of data by HPLC that will only fit in an exponential plot in excel.

If I was to use a linear graph I would use y=mx+c but the data will NOT fit into a linear plot.

Homework Equations

The R2 for the exponential graph = 0.9973

y=0.0044e(0.0017x)

where (0.0017x) is superscript.

The Attempt at a Solution

I have a value for my recovery that I can read off the graph but don't know how to convert this to ug (I can do this if the graph is linear).

I have plotted µg vs peak response.

Thanks!

Hello Newbie_ . Welcome to PF !

Take the natural logarithm, Ln, of both sides of y=0.0044e^(0.0017x) .

If you graph Ln(y) versus x, the graph should be a straight line.

 

[Newbie_]

OK thanks for the info + welcome.

I have plotted concentraion (µg) vs ln response and the line is linear.

However I'm a bit confused about reading results from the graph ie

y=0.0017x-5.4289

My ln of the value I want to read off the graph = -2.57286965




If I did not use ln I would (although the values would not be -ve):

(-2.57286965-(-5.4289))/.0017




How do I calculate given that I have used ln?

ie when do I do the anti-ln? Or is the above calculation correct as I have taken the ln of all standard values + am using the ln of the recovery value?

I hope this makes sense!

Thanks again.

 

[Newbie_]

(-2.57286965-(-5.4289))/.0017




.

I think this is correct as I have just put in a theoretical example and the "right" result came out. I could be wrong though

Thanks

 

[Mentallic]

If you have an exponential function of the form y=Ae^{kx} for some constants A and k, then taking the log of both sides gives \ln(y)=\ln(Ae^{kx})=\ln(A)+\ln(e^{kx})=kx+\ln(A)

So what we have plotted is y'=kx+\ln(A) where y'=\ln(y)

Now, the graph you have in your example is y'=0.0017x-5.4289 so if you want to find the y value of, say, at x=2, then plug x=2 into the equation to get y'=-5.4255 and since y'=\ln(y) then e^{y'}=y thus we get y=e^{-5.4255}=0.00440

If we want it the other way around, that is to find the x value when y=3 for example, we can plug y=3 into the equation and re-arrange to solve for x, or we can re-arrange the equation to make x the subject right off the bat:

\ln(y)=kx+\ln(A)

kx=\ln(y)-\ln(A)

x=\frac{\ln\left(\frac{y}{A}\right)}{k}

So for y=3, A=0.0044, k=0.0017, we have x\approx 3838

 

[NascentOxygen]

It might be clearer if you plotted your data on log-scaled graph paper. You can design and print off a sample at this site: http://incompetech.com/graphpaper/logarithmic/

It can produce both log-linear and log-log. (I thought you'd need log-linear, for e^x to generate a straight line.)