Growth and Decay Student Notes

Math III Name Notes Exponential Growth and Decay Use logarithms to solve exponential growth and decay problems. a) Equation when a quantity increases by a fixed percent: y=a(1+r) (Growth) a is the initial amount. ris the rate of growth (Use for problems such as appreciation, inflation, salary projections, or rate of population growth.) b) Equation when a quantity decreases by a fixed percent: y=a(1-r) (Decay) (Use for problems involving depreciation.) c) Equation often used by scientists: yae", where k is a constant. If k is positive, the function is an exponential growth function. If k is negative, the function is an exponential decay function. (Use for problems such as radioactive decay or half-life.) 1) A computer engineer is hired for a salary of $28,000. She receives a 5% raise each year. a) Find the exponential model. Is it growth or decay? b) After how many years will she be making $50,000 or more? a=28000 r=0.05 y=28000(1+0.05) y=28000(1.05) Growth 50000-28000 (1.05)t 2=1.05 14 log 25 =tlog1.05 t=109 25/4 log 1.05 t=11.88 yrs 2) As technology advances, the price of many technological devices goes down. One brand of hand-held organizer sells for $89. a) Find the exponential model if its price decreases 6% per year. Is it growth or decay? Q=89 y=89(1-0-0u) y= 89 (0.94)t Decay b) How much will it cost after 5 years? ५= 89(0.94)5 y=$65.32 c) After how many years will its price be $50? 50 = 89 (0.94) t t 50=0.94 89 10950/89=tlog0.94 +-log 50/09 1090.94 t=9.3yrs 3) What is the half life of a radioactive isotope if there are 10 grams initially, and there are 6.52 grams left after 1000 years? (Find k first.) Note 1096.529 Decay! -k value kt y=ae 010 t=1000 y=6-52 K(1000) 6.52-108 0-652 e 1000 lno.652=1000k ene K= ln 0.652 1000 K=-0.000428 ② life =5g 10 -0.000478 5-100 =-0.000428+ en½-000042et t= ln½ -0.000428 t=1619.5yrs 4) For Dave to buy a new car comparably equipped to the one he bought years ago would cost $12,500. Since Dave bought the car, the inflation rate for cars like his has been at an average annual rate of 5.1%. a) Find the exponential model. Is it growth or decay? 12500 = a (110051) 12500 = a (1.051) 8 years ago! b) If Dave originally paid $8,400 for the car, how long ago did he buy it? 12500=8400 (1.051) 1251.051 84 10g 25/84 = t log 1.051 t-7.99yrs *5) Cobalt, an element used to make alloys, has several isotopes. One of these, cobalt-60, is radioactive and has a half-life of 5.7 years. Cobalt-60 is used to trace the path of nonradioactive substances in a system. What is the value of k for Cobalt-60? y=aekt a= half life *Decay-k! a=ae5.7k e In 25.7k Ine 5.7 K=-0.1216 6) In 1950, the population of a city was 53,000. Since then, the population has been growing at a rate of 1.3% per year. a=53000 r=0.013 a) Find the exponential model. Is it growth or decay? Growth y = 53000 (1+0.013) y=53000(1.013) b) When will the population be 70,000? 70000=53000(1.013) c) What will the population be in 2030? 4.53000 (1.013)00 79 t 53=1.013 109=tlog1.013 t≈21.5 Y=148,947 1971 7) In a laboratory, a culture increases from 30 to 195 organisms in 5 hours. What is the hourly growth rate? y=a(1+r)t 195=30(1+r)5 =(1+r)5 =+r Y=513 -1 r-0454 45.4% of rabbits in an area is modeled by the growth equation P(1) = 8e026, where P is in thousands and r is in years. How long will it take for the population to reach 25,000? tz en 25/8 0.26 8) The population 25=8e 25 0.210t 0.26t e In2=0.26t ene t=4.38 yrs