### 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 0^{0} 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?

Solution: A=P(1+c/n)^{cn}^{}^{A: 1200 }^{P:600}^{r:0.08}^{n:2}^{t:???}^{}^{1200 = 600(1+0.08/2)2t}^{2=(1.04)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. log10^{x} = 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 loge^{x} = 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.}^{}^{example: }^{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?}^{}^{a.}^{A:6.2}^{A0: 5(tri)t= t= 2001 -1990= 11}^{}^{A=A0(model)t}^{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)}^{(1/11)log(6.2/5)=log(model)}^{0.00849 = log(model)}^{100.00849 = model}^{}^{A=5(100.00849)}^{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}^{}^{b.}^{t = 9 in 1990 - 1999 }^{a(9)= 5(100.00849(9))}^{=5.9622A(9)= 5e0.01956(9)=5.9622Case 2: Given lots of information (A0,M,P)A=Aom(t/p)A: is the amount of "substance " At the end of the time periodA0: 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.Example:A colong of bateria double 6 days. If there were 3000 bacteria to begin with how many bacteria will there be in 15 days?A=A0=3000M=2P=6t=15A=Aomt/pA=3000(2)15/6A=16970.5628There 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 quarterlyhow long will it take you to become a millionaire?.2. It is estimated that 20% of a radioactive substance decays in 30 hourswhat 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?..OOOO SNAP!!!!! XD..Next scriber is daphne}

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