At what time was the yeast added?

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The discussion revolves around a physics problem involving the timing of yeast addition in a murder mystery scenario. Participants calculate the time when the yeast was added to the dough, which triples in volume every four minutes. One user suggests the yeast was added at 3:20, while another calculates it as 3:18, leading to a discussion on solving the problem using a geometric series. The final calculations involve determining the number of terms needed to reach the room's volume capacity and applying logarithmic functions for accuracy. The conversation highlights the importance of precise mathematical methods in solving time-based problems.
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Hi, I had this story in my Physics book

The police was baffled by what seemed to be the prefect nurder of a girlwho had been found, apparently suffocated, in her kitchen. The girl had been making bread in her kitchen, whose dimensions were 6,10, 10 m. she had formed the dough into a ball of volume 1/6 cubic m and turned away to wash some dishes. The criminal added a special virulent strain of yeast to the bread. As a result the bread immediately started to rise in volume, triple every 4 min. before long the dough filled the room, stopping the clock at 3:48 and squashing the girl into the wall. By the time the police came the next day, the yeast worked itself and the dough returned to its normal size.

at what time was the yeast added?

My answer is 3:20 is it true?

M B
 
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I got 3:18. Maybe we rounded differently?
 
I too (like tide) get t ~ 29 min 50 sec
 
Ok!,
Since the answer is not there in my book, I say I solved it as a geometric series, am I right?
 
I am not sure what you mean when you say you "solved it as a geometric series." Please explain. :-)
 
Well,
a geometric series goes like this
a, ar, a(r^2), a(r^3), ...
General term take the following look
ar^(n-1)
now we have the initial volume (1/6 cubic meter) which is the first term, a.
the nth term is the final volume 6*10*10. therefore we need to know how mnay terms are there to reach 600 starting with 1/6
accordingly;
600=(1/6) * 3^(n-1) and solve for n-1
the time taken to reach the volume 600 is (n-1) * 4 min
since it was found that the clock stopped at 3:48
therefore, t0=3:48 - (n-1)*4

M B
 
The question reduces to how you solved that last equation for n. I think it might be clearer if you set up the problem as follows:

Since the volume triples every four minutes you can write the volume as

V = V_0 \times 3^{t/4}

where t is the elapsed time. You can solve for t using logarithms:

t = 4 \frac {\ln V/V_0}{\ln 3}
 
same as my answer without rounding.

Thanks

M B
 

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