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anonymous
 3 years ago
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
anonymous
 3 years ago
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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you want something like \[10e^{rt}\] and you are looking for \(r\) right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ah i see the problem you were using the formula for compounding \(n\) times, but it is continuous compounding

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so i should be using the A= Pe^rt ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so here is what you know when you start you have 10, so you begin with \(10e^{rt}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you also know that when \(t=45\) you have 30, so replace A by 30 and \(t\) by 45 and solve for \(r\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i.e. solve \[30=10e^{45r}\] for \(r\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you good from there? divide by 10 get \[3=e^{45r}\] then take the log and finally divide by 45

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you could have actually started at that step, reasoning that it triple is 45 minutes

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so i would have log3= loge^45r ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thank you so much for explaining this to me! :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no you would have \[\log(3)=45r\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i though what you did to once side you did to the other?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0when you take the log of \(e^{45r}\) you just get \(45r\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh that's right. there are so many rules

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[e^x=y\iff x=\ln(y)\] they say the same thing

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well i wouldn't call that a "rule" it is equivalent logarithmic form or how one gets a variable out of the exponent

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0don'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
 3 years ago
Best ResponseYou've already chosen the best response.0to me they all feel like rules and properties right now, probably until get the hang of it more

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that 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
 3 years ago
Best ResponseYou've already chosen the best response.0hope 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
 3 years ago
Best ResponseYou've already chosen the best response.0you 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
 3 years ago
Best ResponseYou've already chosen the best response.0in a line you needed to find \(m\) in this case you need to find \(r\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that 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
 3 years ago
Best ResponseYou've already chosen the best response.0and 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
 3 years ago
Best ResponseYou've already chosen the best response.0you are so amazing! Thank you for explaining it, makes so much sense!
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