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

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Presentation transcript:

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

Better than investing annually?

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

e ( …): 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.

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

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

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

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?

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!] … = e 0.017t [divide by 603,106] log e = 0.017t [put into log form, and with a base of e] … = 0.017t [evaluate dat log] … = t [divide by 0.017]

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?

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).