Question

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

Answer 1

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

Answer 2

yes.

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

you also know that when \(t=45\) you have 30, so replace A by 30 and \(t\) by 45 and solve for \(r\)

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

so i would have log3= loge^45r ?

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

oh that's right. there are so many rules

Answer 16

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

Answer 17

well i wouldn't call that a "rule" it is equivalent logarithmic form
or how one gets a variable out of the exponent

Answer 18

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

Answer 19

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

Answer 20

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.

Answer 21

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

Answer 22

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 \(t=0\)

Answer 23

yes. i do.

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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