0
$\begingroup$

For example, say you have plotted a standard curve and managed to obtain a logarithimic trendline equation like: y = mln(x) - b

And you are working with data that includes: a) standards,their concentrations and absorbance readings (taken twice - with mean absorbance and blank absorbance also calculated) b)samples (and their absorbance readings) that you want to know the concentration of c) all the samples have been diluted two fold

Would you simply just rearrange the equation y = mln(x) - c to: x = (y + c) / mln
and input y value as being the mean absorbance readings of the samples you want the concentration of?

Are there any further steps after this? Or any I have missed out?

Is there anything I need to do with the dilution factor? Do I have to do mean absorbance - mean blank, to get the correct absorbance readings to input as y or are the mean absorbance readings themselves sufficient?

Thank you

$\endgroup$
1
  • $\begingroup$ You can't write x = (y+c) / mln since ln is not a constant, rather it is a function on x. So, instead write ln(x) = (y + c) / m. $\endgroup$ – Roni Saiba Nov 28 '20 at 7:13
0
$\begingroup$

If by ln you mean the natural logarithm, then no, you can't rearrange the equation in that way. Try this:

$$m\cdot \ln(x) - c = y\\ \ln(x) = y + c\\ \ln(x) = \frac{y+c}{m}\\ \exp(\ln(x)) = \exp\left(\frac{y+c}{m}\right)\\ x = \exp\left(\frac{y+c}{m}\right) $$

$\endgroup$
2
  • $\begingroup$ Yes, ln is for natural logarithm. Thank you, I will try to work it out using this equation. $\endgroup$ – 123 Mar 2 '20 at 11:33
  • 1
    $\begingroup$ @user1136 I'm not sure what new information you've added to this answer, nor do I understand what issue you have with what I have written above. The original question was answered simply by pointing out a mistake in the original poster's algebra. $\endgroup$ – lincolnck Nov 29 '20 at 16:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.