November 08, 2006

Scribe, Problem solving

In morning's class we talked about the equation , how it can't be zero:

Mr.K explain why it can't be zero. Well 00 doesn't not exist. That being zero can

never happen Because it can never touch the x-axis.Also it's undefined.

continuing from last Math dictionary

Example: Gerald invested $600 at 8% interest, compound semi-annually

How long will it take for his investment to double?


A: 1200

1200 = 600(1+0.08/2)2t

ln2 = 2t ln(1.04)
ln2/ln(1.04) =2t
(1/2 )ln2/ln(1.04) =t
8.8365 year = t

it would take 8 years or 10 months

How we got 10 month is:

8.8365 x 12 =10.0379

Common logarithm

The base 10 is used so frequently with logarithms that it is known as the "common logarithm".
when the base of a logarithm is not indicated , base 10 is assumed.

i.e. log10x = logx

Natural logarithm

e = 2.718281828459
e(Euler's number) is a number that a rises naturally in the logarithmic functions;
Partcularly in the case of coutinuous expontential growth ( or Decay)
The natural Log arithm is typicallu written as lnx;read as "Ellen of x"

ie logex = lnx

Exponential modeling

The basic function: f(x) =abx
how we model real life situations depends on what kind, or how much , information we are given:

case 1:

Working with a minimal amount of information(A,Ao, (Tri)t) we will create a model in base 10 and base e... base e is prefered.

A= Ao(model)*

A0: is the original amount of "substance" at the beginning of the time period.
A: is the amount of "substance" as the end of the time period.
Model: is our model for the growth (or decay) of the substance", it is usually an exponential expression in base 10 or base e although any base can be used.
t: is the amount of time that has passed for the substance" to grow(or Decay) from A0 to A.


The population of the earth was 5 billion in 1990. In zool it was 6.2 billion.
a. model the poplution growth using an exponential function.
b.What was the population in 1999?

A0: 5
(tri)t= t= 2001 -1990= 11

6.2 = 5(model)11
The are two ways , but the same answer.

Base 10

Log(6.2/5) = log(model)11

log(6.2/5)= 11 log(model)
0.00849 = log(model)
100.00849 = model

A= 1.0197

Base e

ln(6.2/5)= ln(model)11

ln(6.2/5) = 11 ln(model)
(1/11) ln (6.2/5) = ln(model)
0.01956 = ln(model)
e0.01956 = (model)
A= 5e0.01956t
e= 1.0197

t = 9 in 1990 - 1999
a(9)= 5(100.00849(9))

A(9)= 5e0.01956(9)

Case 2: Given lots of information (A0,M,P)


A: is the amount of "substance " At the end of the time period
A0: is the original amount of "substance" at the beginning of the time period.
M: is the "multiplication factor"or growth rate.
P: is the period; the amount of time required to multiply by "m"once.
t: is the time that has passed.


A colong of bateria double 6 days. If there were 3000 bacteria to begin with how many bacteria will there be in 15 days?


There will be approximately 16971 Bacteria

At the end of the period there were three questions.

1. You have $10,000 and invest it all at 12% compounded quarterly
how long will it take you to become a millionaire?.

2. It is estimated that 20% of a radioactive substance decays in 30 hours
what is the half life of this substance?.

3.The population of toronto was 4000,000 in 2000. The growth is modeled by P(t) = 4000,000 e0.012t what year will the population be 6,400,000?.


Next scriber is daphne

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