I was trying to solve this for two days. how can I find the value of $K$ which is the carrying capacity and the value of a, the equation shows below. $$ \frac{a(1-(26.273/K))}{a(1-(27.165/K))}=\frac{0.03274448}{0.03253040} $$

New contributor
Jessabelle Borja is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
  • 1
    $\begingroup$ Is $a$ a constant? For me, it looks like it can be cancelled away in the equation. Also, please improve the mathematical notation with MathJax so that we know what the exact details in the formula are. You mentioned that you have tried to solve it - what did you try so far? What was the result? $\endgroup$ – Matti P. 2 days ago
  • $\begingroup$ I have cancelled the constant a by ratio&proportion but i was stuck to get the value of K. $\endgroup$ – Jessabelle Borja 2 days ago

After cancelling $a$, you are left with the equation $$ \frac{(1-(26.273/K))}{(1-(27.165/K))}=\frac{0.03274448}{0.03253040} = 1.0065809 $$ In order to remove numerical rounding errors, let's write this as $$ \frac{1-\frac{a}{K}}{1-\frac{b}{K}}= c $$ You can multiply both sides by $1-\frac{b}{K}$ to get $$ 1-\frac{a}{K} = c\left(1-\frac{b}{K} \right) = c - \frac{bc}{K} $$ Let's gather terms with $K$ to the left-hand side and the terms without $K$ to the right-hand side: $$ \frac{bc}{K} - \frac{a}{K} = c-1 $$ $$ \frac{bc-a}{K} = c-1 $$ Continuing, you can multiply both sides by $K$ to get $$ bc-a = (c-1)K $$ and now divide by $c-1$ to get the answer: $$ K = \frac{bc-a}{c-1} $$ Is it clear now?

  • $\begingroup$ i'm still confused. am i going to derive it then? or just distribute the value of the variables? $\endgroup$ – Jessabelle Borja 2 days ago
  • $\begingroup$ If there is anything unclear, please ask. I can help you. $\endgroup$ – Matti P. 2 days ago
  • $\begingroup$ thank you so much. this is a very big help. $\endgroup$ – Jessabelle Borja 2 days ago
  • $\begingroup$ You're welcome. You can mark my answer as accepted, if you feel like it answers your question. $\endgroup$ – Matti P. 2 days ago
  • $\begingroup$ it's done. thanks again. and I hope you will answer if i have questions later on. :) $\endgroup$ – Jessabelle Borja 2 days ago

Your Answer

Jessabelle Borja is a new contributor. Be nice, and check out our Code of Conduct.

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.