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A window through the walls of our classroom. This is an interactive learning ecology for students and parents in my Pre-Cal Math 40S class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.
6 comments:
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.
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
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
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
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