Friday, April 25, 2008

Today's Slides: April 25

Here they are ...



Posted by at
Labels: , ,

6 comments:

zeph said...
This comment has been removed by the author.
April 26, 2008 at 5:21 PM
zeph said...

Seeing that I can't ask questions in class all week, the only way for me to ask questions is to post my questions as a comment on Friday's blog. (The chat box isn't sufficient because there's too much text and it'll be harder to read if I posted my questions there.)

* How do you know which exponential modeling equation to use for the right problem? (In other words, which exponential modeling formula do you use for the right problem?)

* How do you do EXPONENTIAL MODELING I #2 d,e?

2. In 1995, the population of Calgary was 828 500, and was increasing at the rate of 2.2% per year.

a) P = 828500(1.022)^y

d) Write the equation in part (a) as an exponential function with base 2.
Answer: P=828500(2)^(y/32)

e) Write the equation in part (a) as an exponential function with base e.
Answer: P=828500e^(0.022y)

I also don't see how to do #5 of that worksheet.

April 26, 2008 at 5:25 PM
zeph said...
This comment has been removed by the author.
April 26, 2008 at 5:53 PM
zeph said...

How do you do these questions?

EXPONENTIAL & LOGARITHMIC EQUATIONS - REVIEW #15, 16

15. logX^2 = (logX)^2

Answer: 1,100 (Not sure if that's the numbers one (1) and one hundred (100), or one thousand one hundred (1 100) written in American style with the comma.)

16. [log_3(X)]^2 - log_3(x)^2 = 3

Answer: x = 1/3, 27

April 26, 2008 at 5:54 PM
m@rk said...

Zeph,

I'm going to answer your second question first, since it is tougher.

Question 15

logX^2 = (logX)^2

First is to apply the rules for logarithms.The equation will look like this:

2logX= (logX)^2

Now, let logX= y (or some other variable. The equation will look like this:

2y=y^2

Do some algebraic massage

0=2y-y^2

0= y (2-y)

y= 0 and y=2

Now substitute back logX=y

logX=0 and logX=2

Take antilog of both sides

x=10^0 and x=10^2

x=1 and x=100

-m@rk

April 26, 2008 at 11:03 PM
m@rk said...

Zeph,

Now I'm going to show you how to do question 16. Special credits to benofschool for showing me how to do this.

[log_3(X)]^2 - log_3(x)^2 = 3

Apply the rules of logarithms.

[log_3(x)]^2 - 2log_3(x) = 3

Let, log_3(x)= y

y^2-2y=3

Apply algebraic massage

y^2-2y-3=0

Solve for y.

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

y=-1 y=3

Substitute back log_3(x)= y

log_3(x)=-1 and log_3(x)=3

x= 1/3 and 27

I'm going to post the answer for the other questions tomorrow.

-m@rk

April 26, 2008 at 11:15 PM

Post a Comment

Subscribe to: Post Comments (Atom)