Exponential Growth/Decay Review

Learning Targets I can write an exponential equation to model exponential growth/decay. I can write an exponential equation to model an investment or a loan. I can solve an exponential equation for the principal, rate, or amount (given the time).

Equations you should know…   A - Starting Amount r - rate (percent as a decimal) x - time     P – Principal (Initial investment/loan) r – rate (percent as a decimal) n – number of times the interest is compounded per year t – number of years.

y=126.43 Answer: Ticket price will be $126.43 in 2014. Example 1: Exponential Growth The price of Minnesota Vikings tickets were $68.77 in 2005 and has grown 7% per year. What will the ticket prices be in 2014?     Starting Amount A – 68.77 Rate r – 0.07 (7% as a decimal) Time x – 9 years (2014-2005) y – variable we’re solving for Solve for the missing variable.   Enter into calculator. y=126.43 Answer: Ticket price will be $126.43 in 2014.

r = 0.023 Answer: Decaying by 2.3% per hour Example 2: Exponential Decay A petri dish started with 650,000 bacteria. Find the rate the bacteria are decreasing by per minute if there are 228,000 after 45 minutes.     Starting Amount A – 650,000 Rate r – is the variable we solve for Time x – 45 minutes y – 228,000 bacteria after 45 minutes Solve for the missing variable.   Divide both sides by 650,000         Raise both sides to the reciprocal power to undo the exponent.   Subtract 1 on both sides. -1 -1   Divide by -1 on both sides.   r = 0.023 Answer: Decaying by 2.3% per hour

Compound Interest What is Compound Interest? Addition of Interest to an account Note: If interest is compounded monthly, interest is added to your account once a month (12 times a year) Also common… Semi-Annually – 2 times a year Quarterly – 4 times a year Weekly – 52 times a year Daily – 360 times a year

Compound Interest Formula A: amount of money over a given time P: Principal amount (Starting amount) r: YEARLY interest rate (as decimal) N: Frequency of compound t: years

1 48 1 48 Answer: 3% annual interest Example 3: Investment (Solving for the interest rate) You deposit $89 in an account compounded monthly. In 4 years, you have $100.33. Find the annual percent interest. Write an exponential equation to model this investment 100.33=89 1+ 𝑟 12 12∙4 Principle P– 89.00 Rate r – is the variable we solve for Time t – 4 years A – $100.33 (Amount of after 4 years) Compounded N– 12 Solve for the missing variable. 100.33=89 1+ 𝑟 12 12∙4 Divide both sides by 89 89 89 1 48 1 48 0.13 = 1+ 𝑟 12 48 Raise both sides to the reciprocal power to undo the exponent. 1.0025 = 1+ 𝑟 12 Subtract 1 on both sides. -1 -1 0.0025 = 𝑟 12 Multiply by 12 on both sides. ∙12 ∙12 𝑟=0.03 Answer: 3% annual interest

Answer: Approximately $600 invested 6 years ago. Example 4: Investment (Solving for the principle) You have a savings account that is being compounded quarterly with a 4.5% annual interest. The account currently has $785. How much did you invest 6 years ago when the account was opened? Write an exponential equation to model this investment 785=𝑃 1+ 0.045 4 4∙6 Principle P– is the variable we solve for Rate r – 0.045 (4.5% as a decimal) Time t – 6 years A – $785 (Amount of after 6 years) Compounded N– 4 (quarterly) Solve for the missing variable. 785=1.31𝑃 Simplify 1+ 0.045 4 24 =1.31 using calculator. 785=1.31P Divide both sides by 1.31 1.31 1.31 𝑝=600.16 Answer: Approximately $600 invested 6 years ago.

New Material All we are doing that is ‘new’ is getting you to solve for the missing time. However, we have been practicing it for the last few days!!

Example 5: Population (Solving for the time) A town has a population of 500 people. The rate is growing by 15% per year. When will the town double the population. Write an exponential equation to model this investment 1000= 500(1+.15) 𝑥 A – Starting amount (500 people) Rate r – .15 which is 15% Time t – missing Y – ending amount (1000, which is double 500) Solve for the missing variable. 2= (1+.15) 𝑥 Divide by 500 on both sides. log 2 =𝑥 log⁡(1.15) Log both sides and pull the variable out front. log 2 =𝑥 log⁡(1.15) -------- ---------- log(1.15) log(1.15) Answer: 4.95 years