### Scribe Post

hello my fellow classmates! studying hard? hmmmm?

In the morning class we started by going over the answers on yesterday's assignment.

a) 2^x = 3^2(x+5)

ln2^x = ln3^(2x+5)

xln2 = (2x+5) ln3

DO NOT FORGET TO DISTRIBUTE!!!!

xln2 = 2xln3 + 5ln3

(move 2xln3 to the other side)

xln2 - 2xln3 = 5ln3

(we want "x" by it self)

x(ln2 - 2ln3) = 5ln3

(multiply ln2 - 2ln3 on both sides)

x = 5ln3/(ln2 - 2ln3)

b) e^x - 3e^(-x) = 2

(this question is different from the first question, we need to look at this as a quadratic equation)

we must first fix and remove the ugly negative exponent of -3e^(-x) by changing it into this

(-3/e^x) so now the equation will look like this (e^x - 3/e^x - 2 = 0)

( to remove the e^x at the bottom we must multiply everything by (e^x)

(e^x)^2 - 3 - 2e^x = 0 I'm just going to rearrange it (e^x)^2 -2e^x -3 = 0) looks much better isn't it?

Now we can solve this question with no problem at all!

LET (e^x) = r ( i like "r" because its the first letter of my name..^_^)

r^2 - 2r - 3 = 0

(factor it)

(r-3) (r+1) = 0

r=3 r=(-1)

Ln(e^x) = 3 or Ln(e^x) = (-1) *don't forget to check both answers by plugging it in the equation* put a check mark on the answer that is acceptable and put "rejected" on the answer that is wrong. In this question, we reject (x = ln(-1) and accept (x = ln3) be careful when doing this because not all question rejects negative answer sometimes in some questions you have to accept the negative answer and reject the positive answer. be careful mon amis!

d) (logX)^2 = logX^2

(logX)^2 = 2logX *this is similar to question B*

(logx)^2 -2logx = 0

LET logx = t (i like using letter "t" also because its the first letter of my favorite food name tinapay, means soft bread)

t^2 - 2t = 0 ( we can easily factor this)

t(t-2) = 0

t= 0 or t= 2

logx = 0 or logx = 2

(0)^2 = (0)^2 = 1 therefore X= 1

there is no base labelled on the log so it must be in base 10 so (10)^2 = 100 therefore X= 100

for question C: log2^x + log2^(2x-3) = 2 I'm not going to go over this one WHY?! because i know you guys are smarter than the average bear and can do this question in a ~snap

Mr.K then gave us this PROBLEM to solve....

a colony of bacteria grows according to the model: P = Poe^rt

a) if the initial population is 1000 bacteria, what is the value of Po?

b) after 10 days the bacteria population triples. Find r.

c) what will be the bacteria population after 15 days?

~mean while~

Mr.K OMG!!!!! started giving us some goodies!!!! i picked the chocolate bar name Mr.BIG, it was small in package but has BIG taste! yum~yum! it was del-icio-us! We All Thank You For That Lovely Treats Mr.K and Kids!

~back to math~

remember:

Formula: P= Poe^r(t) is the same as A= Poe^r(t) where....

A the amount at the end of the investment period

P the principle (original value)

e = 2.718281828459....

r the percent rate written as a decimal

t the time given

Answers:

a) if the initial population of bacteria is 1000, what is Po?......1000! *that's a hard one*

b) after 10 days, the bacteria will triple in population,Find r.

we know that 1000 is the original value so 1000 times 3 = 3000(A) and the time is 10 days

A = Poe^r(t)

3000 = 1000e^r(10)

(divide everything by 1000)

3 = e^10r

(on this step you can either use log or ln.....i will use ln)

ln3 = (10r)ln e (since lne is = to 1 we can just write 10r)

ln3 = 10r (divide by 10)

1/10 ln3 = r ( use your graphing calculator to find out the value)

and i got r = 0.1086

c) what will be the population after 15 days?

P = 1000e^(0.1086)(15)

= 5196.1524

In the afternoon class, we had our pre-test on logs and exponents, i will put the questions and answers for the people who doesn't have a pre-test paper and will go over some of the answer that i think was hard.

1) log24 = 3log a+ log b then possible values for a and b are? a^3b = 24 a=2 b=3

2) solve for x: ln(5-2x) = w

DON'T DISTRIBUTE ln(5-2x) THAT's A BIG NO NO!

we know that log e^x = ln x so...

5-2x = e^w

(move 5 to the other side)

-2x = e^w - 5

(then divide by -2)

x = -1/2(e^w - 5)

NOW WE CAN SAFELY DISTRIBUTE -1/2(e^w - 5)

x = -e^w/2 + 5/2

3) what is the domain of ln(4-x^2)?

just factor (4 -x^2)

(x-2) (x+2) = 0

x= 2 x= -2

the domain is (-2,2)

4) if a point P(c,1.54) is on the inverse of y = 2^x, then the value of c, correct to four decimal places, must be:

the inverse of y = 2^x is x = 2^y and we know that the Y value is 1.54 so just plug it in...x=2^1.54 is 0.6229

5) the half life of Iodine-131, a radionuclide used in diagnosing thyroid disease, is 8 days.

a) if the X-ray department of the grace hospital has 1000mg of Iodine-131, write the amount of Iodine-131 remaining after t(time) day.

we are given: Ao= 1000 m= 1/2 p=8 A=? t=?

m is the "multiplication Factor"

p is the period; the amount of time required to multiply by "m" once

the formula that I'm going to use is A= Aom^t/p since it gave me "m" and "p"

once we find the values we just need to plug them in the formula. A = 1000(1/2)^t/8

In the morning class we started by going over the answers on yesterday's assignment.

a) 2^x = 3^2(x+5)

ln2^x = ln3^(2x+5)

xln2 = (2x+5) ln3

DO NOT FORGET TO DISTRIBUTE!!!!

xln2 = 2xln3 + 5ln3

(move 2xln3 to the other side)

xln2 - 2xln3 = 5ln3

(we want "x" by it self)

x(ln2 - 2ln3) = 5ln3

(multiply ln2 - 2ln3 on both sides)

x = 5ln3/(ln2 - 2ln3)

b) e^x - 3e^(-x) = 2

(this question is different from the first question, we need to look at this as a quadratic equation)

we must first fix and remove the ugly negative exponent of -3e^(-x) by changing it into this

(-3/e^x) so now the equation will look like this (e^x - 3/e^x - 2 = 0)

( to remove the e^x at the bottom we must multiply everything by (e^x)

(e^x)^2 - 3 - 2e^x = 0 I'm just going to rearrange it (e^x)^2 -2e^x -3 = 0) looks much better isn't it?

Now we can solve this question with no problem at all!

LET (e^x) = r ( i like "r" because its the first letter of my name..^_^)

r^2 - 2r - 3 = 0

(factor it)

(r-3) (r+1) = 0

r=3 r=(-1)

Ln(e^x) = 3 or Ln(e^x) = (-1) *don't forget to check both answers by plugging it in the equation* put a check mark on the answer that is acceptable and put "rejected" on the answer that is wrong. In this question, we reject (x = ln(-1) and accept (x = ln3) be careful when doing this because not all question rejects negative answer sometimes in some questions you have to accept the negative answer and reject the positive answer. be careful mon amis!

d) (logX)^2 = logX^2

(logX)^2 = 2logX *this is similar to question B*

(logx)^2 -2logx = 0

LET logx = t (i like using letter "t" also because its the first letter of my favorite food name tinapay, means soft bread)

t^2 - 2t = 0 ( we can easily factor this)

t(t-2) = 0

t= 0 or t= 2

logx = 0 or logx = 2

(0)^2 = (0)^2 = 1 therefore X= 1

there is no base labelled on the log so it must be in base 10 so (10)^2 = 100 therefore X= 100

for question C: log2^x + log2^(2x-3) = 2 I'm not going to go over this one WHY?! because i know you guys are smarter than the average bear and can do this question in a ~snap

Mr.K then gave us this PROBLEM to solve....

a colony of bacteria grows according to the model: P = Poe^rt

a) if the initial population is 1000 bacteria, what is the value of Po?

b) after 10 days the bacteria population triples. Find r.

c) what will be the bacteria population after 15 days?

~mean while~

Mr.K OMG!!!!! started giving us some goodies!!!! i picked the chocolate bar name Mr.BIG, it was small in package but has BIG taste! yum~yum! it was del-icio-us! We All Thank You For That Lovely Treats Mr.K and Kids!

~back to math~

remember:

Formula: P= Poe^r(t) is the same as A= Poe^r(t) where....

A the amount at the end of the investment period

P the principle (original value)

e = 2.718281828459....

r the percent rate written as a decimal

t the time given

Answers:

a) if the initial population of bacteria is 1000, what is Po?......1000! *that's a hard one*

b) after 10 days, the bacteria will triple in population,Find r.

we know that 1000 is the original value so 1000 times 3 = 3000(A) and the time is 10 days

A = Poe^r(t)

3000 = 1000e^r(10)

(divide everything by 1000)

3 = e^10r

(on this step you can either use log or ln.....i will use ln)

ln3 = (10r)ln e (since lne is = to 1 we can just write 10r)

ln3 = 10r (divide by 10)

1/10 ln3 = r ( use your graphing calculator to find out the value)

and i got r = 0.1086

c) what will be the population after 15 days?

P = 1000e^(0.1086)(15)

= 5196.1524

In the afternoon class, we had our pre-test on logs and exponents, i will put the questions and answers for the people who doesn't have a pre-test paper and will go over some of the answer that i think was hard.

1) log24 = 3log a+ log b then possible values for a and b are? a^3b = 24 a=2 b=3

2) solve for x: ln(5-2x) = w

DON'T DISTRIBUTE ln(5-2x) THAT's A BIG NO NO!

we know that log e^x = ln x so...

5-2x = e^w

(move 5 to the other side)

-2x = e^w - 5

(then divide by -2)

x = -1/2(e^w - 5)

NOW WE CAN SAFELY DISTRIBUTE -1/2(e^w - 5)

x = -e^w/2 + 5/2

3) what is the domain of ln(4-x^2)?

just factor (4 -x^2)

(x-2) (x+2) = 0

x= 2 x= -2

the domain is (-2,2)

4) if a point P(c,1.54) is on the inverse of y = 2^x, then the value of c, correct to four decimal places, must be:

the inverse of y = 2^x is x = 2^y and we know that the Y value is 1.54 so just plug it in...x=2^1.54 is 0.6229

5) the half life of Iodine-131, a radionuclide used in diagnosing thyroid disease, is 8 days.

a) if the X-ray department of the grace hospital has 1000mg of Iodine-131, write the amount of Iodine-131 remaining after t(time) day.

we are given: Ao= 1000 m= 1/2 p=8 A=? t=?

m is the "multiplication Factor"

p is the period; the amount of time required to multiply by "m" once

the formula that I'm going to use is A= Aom^t/p since it gave me "m" and "p"

once we find the values we just need to plug them in the formula. A = 1000(1/2)^t/8

you can see.....this question is already finish! its done! its over man! all you have to do is to show your work and write you final answer.....but I'm not going to........I will leave this to you guys! mwuahahhaahahah( evil laugh) WHY?! because math is not a spectators sport!(evil laugh)!

b) how many days must pass before the Iodine-131 drops to a level of 100mg?

again....this question is pretty much straight forward...remember "A" is the amount of substance at the end of the time period and the question ask you how many days must pass before the final amount of the bacteria drops to 100. you are looking for the time(t)

A = Aom^t/p

100 = 1000(1/2)^t/8

1/10 = (1/2)^t/8

I'm going to use log.

log10^-1 = t/8 log 1/2

multiply by 8

8(-1) = t log 1/2

divide by log 1/2

-8/(log 1/2) = t

~please comment me if i did any question wrong~

jess I CHOOSE YOU to be the next scribe!

~ruschev~out~

## No comments:

## Post a Comment