a yeast grows at a rate proportional to its present size, if the orinally amount doubles in two hours, in how many hours will it triple?
you want t where
2^(t/2) = 3
t/2 log2 = log3
t = log9/log2 = 3.17
Ah, yeast, the tiny organisms that make bread rise and give beer its bubbly goodness. So, according to my calculations, if the yeast doubles in size in two hours, we can assume that it's growing at a rate of 1 doubling per 2 hours.
Now, to determine how long it will take to triple, we can apply some clownish logic. Since doubling and tripling are both based on the original size, it will take approximately 3 hours for the yeast to triple. But hey, yeast can be quite unpredictable, so consider this answer with a grain of salt, or should I say, a pinch of yeast?
To solve this problem, we can create a differential equation that represents the growth of the yeast. Let "Q" represent the amount of yeast at any given time "t".
The differential equation can be written as follows:
dQ/dt = kQ
Where "k" is the constant of proportionality.
We are given that the amount of yeast doubles in two hours. This means that when t = 2, Q = 2Q0 (where Q0 is the initial amount of yeast).
Substituting these values into the differential equation:
k(2Q0) = d(2Q0)/dt
2kQ0 = 2Q0/dt
Canceling out the common terms:
2k = 1/2
Simplifying:
k = 1/4
Now we can solve for the time "t" it takes for the yeast to triple. We know that when Q = 3Q0:
dQ/dt = kQ
dQ = kQ dt
Integrating both sides:
∫ dQ/Q = ∫ k dt
ln(Q) = kt + C
ln(3Q0) = (1/4)t + C
Taking exponential of both sides:
3Q0 = e^[(1/4)t + C]
Since C is the constant of integration, we can rewrite it as e^C:
3Q0 = e^(1/4)t * e^C
Combining constants:
3Q0 = C1 * e^(1/4)t
Let C2 = 3Q0/C1:
3Q0 = C2 * e^(1/4)t
Dividing by Q0:
3 = C2 * e^(1/4)t
Now we can solve for "t". Let's isolate the exponential term by dividing both sides by C2:
3/C2 = e^(1/4)t
Taking the natural logarithm of both sides:
ln(3/C2) = (1/4)t
Simplifying:
t = (4/1) * ln(3/C2)
Finally, we substitute C2 = 3Q0/C1:
t = (4/1) * ln(3 / (3Q0/C1))
t = (4/1) * ln(C1)
Therefore, the time it takes for the yeast to triple is given by (4/1) * ln(C1), where C1 is the constant of integration.
To determine the time it takes for the yeast to triple in amount, we need to set up a proportion based on the given information.
Let's assume the original amount of yeast is denoted by "A" (we don't know the specific quantity) and the time it takes for it to double is 2 hours. This means that after 2 hours, the amount of yeast will be 2A.
Now, we want to find out how long it takes for the yeast to triple in amount. Let's represent the time it takes for it to triple as "t" (we don't know this value yet).
Since the growth rate is proportional to the present size, we can set up the following proportion:
2A (amount after 2 hours) / A (original amount) = t (time to triple) / 2 hours (given time for doubling)
We can simplify this proportion by cross-multiplying:
2A * 2 hours = A * t
4A hours = A * t
Simplifying further, we can divide both sides of the equation by A:
4 hours = t
Therefore, it will take 4 hours for the yeast to triple in amount.