9514 1404 393
Answer:
7 days
Step-by-step explanation:
(a) Filling in the given information will result in two equations in the two unknowns. Dividing one by the other will tell you the value of b.
n=2
1.8 = A₀·b²
n=3
2.4 = A₀·b³
Dividing the second equation by the first gives ...
2.4/1.8 = (A₀·b³)/(A₀·b²) = b
4/3 = b
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(b) Substituting this into the first equation, we can find A₀.
1.8 = A₀·(4/3)²
1.8(9/16) = A₀ = 1.0125
A₀ is the value of A when n=0. It is the area of the patch at the beginning of observation, before any days passed.
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(c) The area will be 7 cm when ...
1.0125·(4/3)^n = 7
(4/3)^n = 7/1.0125 . . . . . . . . . . . . divide by A₀
n·log(4/3) = log(7/1.0125) . . . . . . take logarithms
n = log(7/1.0125)/log(4/3) ≈ 6.72092
It will take about 7 days for the area of the patch to exceed 7 cm.
