- 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

- chestercat

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- anonymous

you want something like
\[10e^{rt}\] and you are looking for \(r\) right?

- anonymous

yes.

- anonymous

ah i see the problem
you were using the formula for compounding \(n\) times, but it is continuous compounding

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## More answers

- anonymous

so i should be using the A= Pe^rt ?

- anonymous

so here is what you know
when you start you have 10, so you begin with \(10e^{rt}\)

- 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

i.e. solve
\[30=10e^{45r}\] for \(r\)

- anonymous

you good from there?
divide by 10 get
\[3=e^{45r}\] then take the log and finally divide by 45

- anonymous

you could have actually started at that step, reasoning that it triple is 45 minutes

- anonymous

so i would have log3= loge^45r ?

- anonymous

thank you so much for explaining this to me! :)

- anonymous

no you would have
\[\log(3)=45r\]

- anonymous

i though what you did to once side you did to the other?

- anonymous

when you take the log of \(e^{45r}\) you just get \(45r\)

- anonymous

oh that's right. there are so many rules

- anonymous

\[e^x=y\iff x=\ln(y)\] they say the same thing

- 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

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

to me they all feel like rules and properties right now, probably until get the hang of it more

- 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

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

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

yes. i do.

- anonymous

in a line you needed to find
\(m\)
in this case you need to find \(r\)

- 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

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

you are so amazing! Thank you for explaining it, makes so much sense!

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