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

1. Apr 11, 2012

### Newbie_

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

The R2 for the exponential graph = 0.9973

y=0.0044e(0.0017x)

where (0.0017x) is superscript.

3. 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!

2. Apr 11, 2012

### SammyS

Staff Emeritus
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.

3. Apr 11, 2012

### 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.

Last edited: Apr 11, 2012
4. Apr 11, 2012

### Newbie_

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

5. Apr 11, 2012

### 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 $x\approx 3838$

6. Apr 11, 2012

### Staff: Mentor

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 ex to generate a straight line.)