The Special Number - e

Suppose that f(n) = ( (n+1)/n )n

Then f(1) = (2/1)1 = 21 = 2

f(2) = (3/2)2 = (1.5)2 = 2.25

f(3) = (4/3)3 = (1.33..)3 = 2.370…

f(10) = (11/10)10 = (1.1)10 = 2.5937…

f(100) = (101/100)100 = (1.01)100 = 2.7048…

f(1000) = (1001/1000)1000 = (1.001)1000 = 2.7169…

f(1000000) = (1000001/1000000)1000000 = (1.000001)1000000 = 2.7183…

f(100000) = (100001/100000)100000 = (1.00001)100000 = 2.7183…

As n f(n) 2.7183….

This value is given the name e !

Many real life growth and decay functions can be modelled mathematically using exponential functions with base e.

This can be found on your calculator either by itself as

ex or above the Inx key.

Example1 e10 = 22026.5

e-0.3 = 0.741 etc

Example2 Solve ex = 10 to 2 decimal places.

********Using trial & error

n = 2 e2 = 7.39 too small

n = 3 e3 = 20.09 too large

n = 2.3 e2.3 = 9.97 too small

n = 2.4 e2.4 = 11.02 too large

n = 2.32 e2.32 = 10.18 too large

n = 2.31 e2.31 = 10.07 too large

(***)

To 2 dec places if ex = 10 then x = 2.30

Example3

A radioactive substance decays according to the formula

At = A0e(-2.1t)

where A0 is the original amount and At is the amount remaining after t hours.

(a) Find how much of 530g remains after 31/2 hours.

(b) Show that the half-life is 20 mins !

********

(a) At = A0e(-2.1t)

= 530 X e(-2.1 X 3.5)

= 530 X e(-7.35)

= 0.34g

(b) After one half-life amount left = 530g 2 = 265g

If t = 20mins = 1/3hr A(1/3) = 530 x e(-2.1 3)

= 530 x e(-0.7)

= 263g (sufficiently close)

ALTERNATIVELY

We need At = 0.5A0 and At = A0e(-2.1t)

So A0e(-2.1t) = 0.5A0

So e(-2.1t) = 0.5

Taking t = 1/3 we get e(-2.1 3) = 0.4966

Again sufficiently close to 0.5 so half-life is 20 mins.