Here's the question you clicked on:
confusedstudent2012
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
you want something like \[10e^{rt}\] and you are looking for \(r\) right?
ah i see the problem you were using the formula for compounding \(n\) times, but it is continuous compounding
so i should be using the A= Pe^rt ?
so here is what you know when you start you have 10, so you begin with \(10e^{rt}\)
you also know that when \(t=45\) you have 30, so replace A by 30 and \(t\) by 45 and solve for \(r\)
i.e. solve \[30=10e^{45r}\] for \(r\)
you good from there? divide by 10 get \[3=e^{45r}\] then take the log and finally divide by 45
you could have actually started at that step, reasoning that it triple is 45 minutes
so i would have log3= loge^45r ?
thank you so much for explaining this to me! :)
no you would have \[\log(3)=45r\]
i though what you did to once side you did to the other?
when you take the log of \(e^{45r}\) you just get \(45r\)
oh that's right. there are so many rules
\[e^x=y\iff x=\ln(y)\] they say the same thing
well i wouldn't call that a "rule" it is equivalent logarithmic form or how one gets a variable out of the exponent
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
to me they all feel like rules and properties right now, probably until get the hang of it more
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.
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
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 \(t=0\)
in a line you needed to find \(m\) in this case you need to find \(r\)
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
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
you are so amazing! Thank you for explaining it, makes so much sense!