APPLICATIONS OF THE EXPONENTIAL AND THE LOGARITHM Session 17/citytech/ precalculus APPLICATIONS OF THE EXPONENTIAL AND THE LOGARITHM

f(0) = 5, f(1) = 20 b) f(0) = 3, f(4) = 48 c) f(2) = 160, f(7) = 5 15.1. Let f(x) = c · b^x Determine the constant c and base b under the given conditions f(0) = 5, f(1) = 20 b) f(0) = 3, f(4) = 48 c) f(2) = 160, f(7) = 5 d) f(−2) = 55, f(1) = 7

Solutions:

y(t) = 2·1.02^t y(4) = 2 · 1.02^4 ≈ 2.16. log(5)/ log(1.02) = t≈ 81.3 15.2. The mass of a bacteria sample is 2 · 1.02^t grams after t hours. a) What is the mass of the bacteria sample after 4 hours? b) When will the mass reach 10 grams? y(t) = 2·1.02^t y(4) = 2 · 1.02^4 ≈ 2.16. log(5)/ log(1.02) = t≈ 81.3

15. 3. The population size of a country was 12 15.3. The population size of a country was 12.7 million in the year 2000, and 14.3 million in the year 2010. a) Assuming an exponential growth for the population size, find the formula for the population depending on the year t. b) What will the population size be in the year 2015, assuming the formula holds until then? c) When will the population reach 18 million?

Solution

Solution: y(t) = c · b^t a) 14. 3 = 12. 7 · b^10 b =(14. 3/12

b) We calculate the population size in the year 2015 by setting t = 15: y(15) = 12.7 · 1.012^15 ≈ 15.2

c)We seek t so that y(t) = 18. We solve for t using the logarithm. Divide by 12.7 and than find the log of both sides t ≈ 29.2

15.4. An exponential function with a rate of growth r is a function f(x) = c · b ^ x with base b = 1 + r .

15.6. The number of PCs that are sold in the U.S. in the year 2011 is approximately 350 million with a rate of growth of 3.6% per year. Assuming the rate stays constant over the next years, how many PCs will be sold in the year 2015?

NUMBER OF SALES IN THE YEAR 2015 we set t = 0 for the year 2011, we find that c = 350, Since the rate of growth is r = 3.6% = 0.036 b = 1 + r = 1.036, y(t) = 350 · 1.036^t y(4) = 350 · 1.0364 ≈ 403 NUMBER OF SALES IN THE YEAR 2015