Determine the doubling time or half-life of y in each of the following exponenti

Question

Determine the doubling time or half-life of y in each of the following exponential equations. You may imagine that the variable y represents a quantity like population or light intensity that may be increasing or decreasing as T gets larger, and the variable T represents time or depth or something similar. If y increases as T gets larger look for a doubling time. If y decreases as T gets larger look for a half-life. Your answers should be positive numbers.

y=A × 23^T where A0. (You'll find it doesn't matter what A is as long as A0.)

b) y=B × (12)^5T where B0. (It doesn't matter what B is as long as B0.)

c) y=0.35 × 46^T .

d) y=30 × 0.127^ T.

e) y=0.001× 0.350.01^T

Explanation / Answer

Solution:

a) y(0) = A.
A × 2^(3T) = 2A {because 2 > 1}
2^(3T) = 2
3T = 1
T = 1/3

b) y(0) = B
B × (½)^(7T) = (½)B {because ½ < 1}
(½)^(7T) = (½)
7T = 1
T = 1/7

c) y(0) = .36
.36 × 4^(5T) = 2 × .36 {because 4 > 1}
4^(5T) = 2
2^(10T) = 2
10T = 1
T = 1/10

d) y(0) = 30
30 × .12^(6T) = 15 {because .12 < 1}
.12^(6T) = ½
6T ln (.12) = ln (½)
6T = ln(½) ÷ ln(.12) = -.69315 ÷ -2.12026 (rounded) = .32692 (rounded)
T = 0.05449

e) y(0) = .001
.001 × .36^(0.01T) = .0005 {because .36 < 1}
.36^(0.01T) = ½
.01T ln(.001) = ln(½)
.01T = ln(½) ÷ ln(.001) = -.69315 ÷ -6.90776 (rounded) = .10034 (rounded)
T = 10.034

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