SOLUTION: Climate Change The most abundant greenhouse gas is carbon dioxide. According to figures from the Intergovernmental Panel on Climate Change (IPCC), the amount of carbon dioxide in

Algebra ->  Exponents -> SOLUTION: Climate Change The most abundant greenhouse gas is carbon dioxide. According to figures from the Intergovernmental Panel on Climate Change (IPCC), the amount of carbon dioxide in      Log On


   



Question 1197444: Climate Change The most abundant greenhouse gas is carbon dioxide. According to figures from the Intergovernmental Panel on Climate Change (IPCC), the amount of carbon dioxide in the atmosphere (in parts of volume per million) can be approximated by
C(t) ≈ 280e0.00127t parts per million
where t is time in years since 1750.†
(a)
Use the model to estimate the amount of carbon dioxide in the atmosphere in
1975, 2000, 2050, and 2100.
(Round your answers to the nearest whole number.)
year 1975 2000 2050 2100

C(t) parts per million

(b)
According to the model, in what year, to the nearest decade, will the level surpass 380 parts per million?

Answer by math_tutor2020(3816) About Me  (Show Source):
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Part (a)

t is the number of years since 1750
t = 0 represents 1750
t = 1 represents 1751
and so on

The year 1975 means t = 1975-1750 = 225
The gap from 1750 to 1975 is 225 years
Use this idea for the other years of 2000, 2050, and 2100.

Use spreadsheet software to quickly compute the outputs.
You'll need to type in the exp function instead of something like e^x
Example calculation: =280*EXP(0.00127*225) yields the approximate result of 372.612722644863 which rounds to 373.

This is what you should get when rounding each C(t) value to the nearest whole number.
Yeart = Number of years since 1750C(t) = Amount of CO2 (ppm)
1975225373
2000250385
2050300410
2100350437
ppm = parts per million

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Part (b)

The table above shows
C(225) = 373 .... year 1975
C(250) = 385 .... year 2000

The level surpasses 380 ppm between the years 1975 and 2000.
It's likely it's somewhere close to the year 2000 compared to the year 1975.

Use a natural logarithm to determine a more accurate time value.
C(t) = 280*e^(0.00127t)
380 = 280*e^(0.00127t)
380/280 = e^(0.00127t)
0.00127t = Ln(380/280)
t = Ln(380/280)/0.00127
t = 240.457991772584

Rounding to the nearest year gets us t = 240
240 years after 1750 is 1750+240 = 1990 which also is the start of a decade, meaning we don't have to round to the nearest decade.

Answer: 1990