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Tuesday, May 30, 2006

Rap-up of Exponents and Logarithm

My last scribe ever...YAY! I'm pretty happy about that but I'm also disappointed that my own right hand man *ahem Jefferson betrayed me! So much for taking over the blog. Anyways I guess I'm on my own now and I should get right to it since we did so much work today, so here it goes...

In the morning class we got right down to business. Mr. K put up old exam questions on the board. There were two parts, a calculator and a non calculator part. The first part you were allowed to use a calcualtor:

1) 3x+4 = 72x+1
2) If loga2=0.3562 and loga5=0.8271 show that loga40=1.8957
3) A new car costs $24 000. It's value after t years is given by V=24000(0.8)t
a.) Determine the value after 8 years.
b.) How many years will it take for its value to decrease to one-eighth of its initial value?

For the rest of the questions calculators weren't allowed:

4) A population triples every 7 days. Determine an expression (formula) for the total population after t days if the initial population is 800.
5) Evalute: log(100 sin(π/2))
6) State the range of f(x)=2-x
7) Given:

1/2 ≤ log x ≤ 2

find the minimal and maximum value of x.

Solutions:

1)      log3x+4 = log72x+1          ← change it as a log
(x+4)log3 = (2x+1)log7 ← by the rules of logarithm
xlog3+4log3 = 2xlog7+log7 ← distribute the brackets because the brackets are #'s
xlog3-2xlog7 = log7-4log3 ← get all the #'s and x's together
x(log3-2xlog7) = log7-4log3 ← factor out the x
x = (log7-4log3) ← solve for x using a calculator
(log3-2log7)
x = 0.8766

2) loga40 = loga(23∙5)
= loga23+loga5
= 3loga2+loga5
= 3(0.3562)+(0.8271)
= 1.0686+0.8271
= 1.8957


3.a) V(8)=24000(0.8)8
= 4026.5318
3.b) (1/8)24000 = 3000
original value=24000 final=3000 therefore,

3000 = 24000(0.8)t
24000 24000
(1/8) = (0.8)t
ln(1/8)=tln(0.8)
ln(1/8)=t
ln(0.8)
9.3189 ≈ t ← using a calculator evaluate t


4) P =7days           A=A°Mt/p
M =3 A=800(3)t/7 ← just substitute the values
A° =800
t =?

5) log(100 sin(π/2))
=log(100∙1) ← since sin(π/2)=1
=log(100)
=2

6) f(x)=2-x
So to solve this question we must think back on transformations.
We know that this is an exponential decay because the basic function (2x) is an
exponential growth. Now in transformations when x is negative that means it "flips
over the y-axis" in other words the x-coordinates switch.
Therefore the answer must be (0,∞) because it only approaches 0 but never touches it.

7) 101/2 ≤ 10logx ≤ 102 ← changing it to base 10 will simplify it
√10 ≤ x ≤ 100


Everyone was kind of stuck on that last question so Mr. K
tried to clear it up by giving us another question which
he thought was mildy amusing.
log(logx)=2
102=logx
100=logx
10100=x which we know as Googol!
humorous? i didnt think so.


Now that took the whole morning class and in the afternoon class we wrote in our dictionary for the whole period. There were five pages of notes so for those students that missed class or didn't finish copying here it is:

PROOFS OF THE LAWS OF LOGARITHMS

Proof of the the product laws
Prove logaMN = logaM + logaN
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
x=logaM IFF ax=M
y=logaN IFF ay=N

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
logaMN=loga(ax∙ay)
=logaa(x+y)
=x+y
=logaM+logaN
Q.E.D


Proof of the quotient law
Prove loga(M/N)=logaM-logaN
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
x=logaM IFF ax=M
y=logaN IFF ay=N

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
loga(M/N)=loga(ax/ay)
=logaa(x-y)
=x-y
=logaM-logaN
Q.E.D


Proof of the power law
Prove logaMc=clogaM
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
logaM=b IFF ab=M

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
logaMc=loga(ab)c
=logaacb
=cb
=clogaM
Q.E.D


Proof of the base law
Prove:
logaMc= logbM
logba
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
logaM=c IFF ac=M

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
ac=m
logbac=logbM
clogba= logbM
logba
c=logbM
logba
logaM= logbM
logba
Q.E.D


EXPONENTIAL MODELING

the basic function:

How we model a real life situation depends on how much information we are given.

case 1: Given a minimal amount of information (A, A°, ▲t) we will create a model in base 10 or base e. (base e is prefered)
A is the amount at the end of the time period.
A° is the original amount of the "substance".
MODEL is an exponential function that describes the growth (or decay) of the "substance".
t is the time required for the amount of the "substance" to change from A° to A.

NOTE: Some "substances" are populations. The model is usually an exponential function of the form 10k, or er where k and r are constants.

EXAMPLE: The population of earth was 5 billion in 1990. In 2003 the population grew to 6.3
a) Model the population as an exponential function.
b) What was the population of earth in 1967?


a)                let t=o in 1990.
A°=5 A=A°(model)t
A=603 (in 2003) 6.3=5(model)13
t=2003-1990 6.3 =(model)13
=13 5
From here you can go on in two different ways. It wouldn't really matter which way you choose because it will end up having the same answer.

if you choose base 10                            if you choose base e
log(6.3/5)=log(model)13 ln(6.3/5)=ln(model)13
log(6.3/5)=13log(model) ln(6.3/5)=13ln(model)
1/13log(6.3/5)=log(model) 1/13ln(6.3/5)=ln(model)
0.0077=log(model) 0.0178=ln(model)
100.0077= model e0.0178= model
therefore P=5(100.0077t) P=5e0.0178t

b) If t=0 in 1990 then t=-23 in 1967.
if base 10 if base e
P=5(100.0077(-23) P=5e0.0178(-23)
P=3.3219 P=3.3219


case 2: Given lots of information (A, A°, M, p)A is the amount at the end of the time period t.
A° is the original amount.
M is the "multiplication factor" or growth rate.
t is the time has passed.
p is the time period; time required to multiply by M once.

EXAMPLE 1: A colony of bacteria double every 6 days. If there were 3000 bacteria at the start of an experiment how many bacteria will there be in 15 days?
 A=?             A=A°Mt/p
A°=3000 =3000(2)15/6
M=2 =16970.5628
t=15 ≈16971 bacteria
p=6


EXAMPLE 2: The half life of carbon 14 is 5700 years. How much of a 10mg sample will remain after 4500 years?
 A=?             A=A°Mt/p
A°=10 =10(1/2)4500/5700
M=1/2 =5.8022 mg
t=4500
p=5700
And that's that!! Phew...three five small coffee cups later and I'M FINALLY DONE! YAY! Okay so now that I'm all finished and I'm in a cranky mood I will pick none other than michael as the scribe for tomorrow. Nah, enjoy!



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4 Comments:

At 5/31/2006 7:28 AM, Blogger Mr. H said...

Nice Scribe Janet. You are a master of using colour and text size to emphasize the needed points. This post deserves to be in the Hall of Fame.

Thanks for putting so much effort into explaining the rap up on Exponents and Logarithms.

Mr. Harbeck
Sargent Park School

 
At 5/31/2006 9:23 AM, Blogger Mr. Kuropatwa said...

WOW! You've been inducted into the Scribe Post Hall Of Fame!

Great use of the <pre> tag. Unfortunately it caused some of the text to spill over into the side bar. If you fix up the formating it will look perfect. ;-)

 
At 5/31/2006 8:27 PM, Blogger mark said...

grats JAnet. really nice scribe. oh and mr.k michael wasn't here today so does his scribe move for tomorow?

niceee janet! =D

 
At 5/31/2006 9:13 PM, Blogger Mr. Kuropatwa said...

Probably, bust since Manny covered for Mike I guess it's Manny's call.

 

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