anonymous
  • anonymous
You are growing yeast for a science experiment. You start with 10 ml. You set the experiment up, go away and return 45 minutes later. When you come back, your have 30 ml. Find the rate at which it is growing. The EXACT answer (no calculator necessary) can be given in the form a  ln(b). So I do know the answer to this, I got it wrong on the test i took last week, but I am hoping someone can explain to me the process of how to get the answer. I had used to A=(1+r/n)^(nt) but i ended up getting so lost in the problem that I left it blank
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
you want something like \[10e^{rt}\] and you are looking for \(r\) right?
anonymous
  • anonymous
yes.
anonymous
  • anonymous
ah i see the problem you were using the formula for compounding \(n\) times, but it is continuous compounding

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
so i should be using the A= Pe^rt ?
anonymous
  • anonymous
so here is what you know when you start you have 10, so you begin with \(10e^{rt}\)
anonymous
  • anonymous
you also know that when \(t=45\) you have 30, so replace A by 30 and \(t\) by 45 and solve for \(r\)
anonymous
  • anonymous
i.e. solve \[30=10e^{45r}\] for \(r\)
anonymous
  • anonymous
you good from there? divide by 10 get \[3=e^{45r}\] then take the log and finally divide by 45
anonymous
  • anonymous
you could have actually started at that step, reasoning that it triple is 45 minutes
anonymous
  • anonymous
so i would have log3= loge^45r ?
anonymous
  • anonymous
thank you so much for explaining this to me! :)
anonymous
  • anonymous
no you would have \[\log(3)=45r\]
anonymous
  • anonymous
i though what you did to once side you did to the other?
anonymous
  • anonymous
when you take the log of \(e^{45r}\) you just get \(45r\)
anonymous
  • anonymous
oh that's right. there are so many rules
anonymous
  • anonymous
\[e^x=y\iff x=\ln(y)\] they say the same thing
anonymous
  • anonymous
well i wouldn't call that a "rule" it is equivalent logarithmic form or how one gets a variable out of the exponent
anonymous
  • anonymous
don't forget you are trying to solve for a variable that is in the sky (exponent) algebra will not do that for you you need logs for that
anonymous
  • anonymous
to me they all feel like rules and properties right now, probably until get the hang of it more
anonymous
  • anonymous
that makes sense. thank you so much! i hate when i get a problem wrong on a test and i can't figure out where i went wrong.
anonymous
  • anonymous
hope it is clear though. you are looking for \(r\) and it is an exponent that is why this is different from algebra. you need to take the log to get it
anonymous
  • anonymous
you remember finding the equation of a line given two points? this is analogous to that \(b\) is the \(y\) intercept, what you get when \(x=0\) in this case \(P\) is the \(y\) intercept, what you get when \(t=0\)
anonymous
  • anonymous
yes. i do.
anonymous
  • anonymous
in a line you needed to find \(m\) in this case you need to find \(r\)
anonymous
  • anonymous
that actaully makes so much sense to me. i am always trying to bring the r and t down like numerical exponents and solving them algebraically. .. which is usually wrong
anonymous
  • anonymous
and you do it almost like you do with a line if you have \(y=4x+b\) and you did not know \(b\) you could find it if say \((3,5)\) is on the graph by saying \[5=4\times 3+b\] and solving for \(b\) this is almost the same we know \((45,30)\) is on the graph so you can solve \[30=10e^{45r}\] for \(r\) using the log
anonymous
  • anonymous
you are so amazing! Thank you for explaining it, makes so much sense!

Looking for something else?

Not the answer you are looking for? Search for more explanations.