The island of Slashenburn is depleting its rainforests at an alarming rate. In 1960 the forests were untouched, however by 1979 only 92 % remained. Assuming that the amount remaining is decreasing exponentially, determine: a) What percentage will remain in the year 2008?
Question
The island of Slashenburn is depleting its rainforests at an alarming rate. In 1960 the forests were untouched, however by 1979 only 92 % remained. Assuming that the amount remaining is decreasing exponentially, determine: a) What percentage will remain in the year 2008?
Solution
To solve this problem, we need to use the formula for exponential decay, which is:
A = P * e^(kt)
where: A is the amount remaining after time t, P is the initial amount (in this case, 100%), k is the decay constant, t is the time elapsed (in years), and e is the base of the natural logarithm (approximately 2.71828).
First, we need to find the decay constant, k. We know that in 1979 (19 years after 1960), 92% of the forest remained. So we can set up the equation:
92 = 100 * e^(19k)
To solve for k, we first divide both sides by 100:
0.92 = e^(19k)
Then we take the natural logarithm of both sides:
ln(0.92) = 19k
Solving for k gives:
k = ln(0.92) / 19 ≈ -0.00436
Now that we have k, we can find out what percentage of the forest will remain in 2008. This is 48 years after 1960, so we set up the equation:
A = 100 * e^(48*-0.00436)
Solving this gives:
A ≈ 100 * e^(-0.20928) ≈ 81.1%
So, approximately 81.1% of the forest will remain in 2008, assuming the rate of depletion continues exponentially at the same rate.
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