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Suppose Marcello invests $500 at 1.2% annually. How long will it take for that amount to double?

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Presentation on theme: "Suppose Marcello invests $500 at 1.2% annually. How long will it take for that amount to double?"— Presentation transcript:

1 Suppose Marcello invests $500 at 1.2% annually. How long will it take for that amount to double?

2 Better than investing annually?

3 e (2.71828…): the natural base It represents the base rate of growth shared by all continuously growing processes.

4 e (2.71828…): the natural base Q: Why the natural base? A: Because it shows up in population growth, radioactive decay, and in systems that exhibit continuous growth or decay.

5 Suppose we applied a (theoretical) continuous growth rate to an investment. Continuously Compounding Interest Formula: A = Pe rt

6 A = Pe rt Note that the expression has been substituted with e r now.

7 Take Tacoma’s population (as of 2012) of 202,010 which was growing at 1.8%. If the growth rate remains the same, what will its population be in 2013? Continuously Compounding Interest Formula: A = Pe rt = 202,010e 0.018*3 = ~213218

8 It’s e-asy! If Portland, with its 2012 population of 603,106, is continuously growing at 1.7% then what will its population be in 2017?

9 How e-nteresting Portland, with its 2012 population of 603,106, is continuously growing at 1.7%. When what will its population surpass 700,000? 700,000 = 603,106e 0.017t [solving for t!] 1.16065… = e 0.017t [divide by 603,106] log e 1.16065 = 0.017t [put into log form, and with a base of e] 0.148987… = 0.017t [evaluate dat log] 8.76396… = t [divide by 0.017]

10 e-nsane in the Membrane Portland, with its 2012 population of 603,106, is continuously growing at 1.7%. When what will its population surpass 800,000?

11 oh btw log e x = ln(x) The natural base deserves a natural logarithm. So from now on, instead of using log e (3), a shortcut to use is now ln(3).


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