Jane has been growing two bacteria farms. Bacteria farm Rod has a starting population of 2 bacteria, while Bacteria farm Sphere has a starting population of 8 bacteria. However, Jane starts growing Rod five hours before she starts growing Sphere. At 8 p.m., Jane checks her farms and finds that they have the exact same population. If the population of Rod doubles every hour, but the population of Sphere is quadrupled every hour, how many hours ago did she start growing Sphere?

Answers

Answer 1
Answer:

Answer: Jane started growing Sphere 3 hours ago

Step-by-step explanation:

Farm Rod starting population (Rsp) = 2

Farm Sphere starting population (Ssp) = 8

Let´s name "Rh" the quantity of hours since Rod started growing, and

"Sh" the quantity of hours since Sphere started growing.

And, let´s name "R" the population of farm Rod at 8 p.m. and "S" the population of farm Sphere at 8 p.m.

Population of Rod doubles every hour, therefore:

R = Rsp * 2^(Rh)

R = 2(2^(Rh))

Population of Sphere is quadrupled every hour, therefore:

S = Ssp * 4^(Rh)

S = 8(4^(Rh))

At 8 p.m. Jane found that R = S

Therefore, at 8 p.m:

2(2^(Rh)) = 8(4^(Sh))

dividing both sides by 2

2^(Rh) =4(4^(Sh))

adding exponents

2^(Rh) = 4^(Sh+1)

2^(Rh) =2^{2^(Sh+1) }

the bases are the same; exponents must be the same

Rh = 2Sh + 2    (equation 1)

And we also know that Jane started growing Rod five hours before Sphere:

Rh = Sh + 5    (equation 2)

Replacing equation 2 into equation 1:

(Sh + 5) = 2Sh + 2

5 - 2 = 2Sh - Sh

3 = Sh, or

Sh = 3

Jane started growing Sphere 3 hours ago.  


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