October 27, 2006

Scribe Post

The morning was nothing new since we had a test the entire period, so there was one or two new bits of information during the afternoon class.

When we first came in the usual questions were on the board.

Solve for x:
8^x=32
2^3x=2^5
3x=5
x=5/3
3-3^x-1=1/27
3^1+(x-1)=3^-3
1+x-1=-3
x=-3
5x^4=80
x^4=16
(x^4)^1/4=(16)^1/4
x=2
1/4^x-2=64
4^-x+2=4^3
-x+2=3
-x=1
x=-1
2(5^2x-9)=250
5^2x-9=125
5^2x-9=5^3
2x-9=3
2x=12
x=6

There was one last question which gave people plenty of problems since almost everyone was apparently making the same mistake in the solving.

2^2x=3(2^x)-2
(2^x)^2-3(2^x)+2=0
LET a = 2x
a^2-3a+2=0
(a-1)(a-2)=0
a=1 or a=2
~~~~~~~~~~~~~~~~~~~
2^x=1
2^2=2^0
x=0
~~~~~~~~~~~~~~~~~~~
2^x=2
x=1

The common mistake that people were making was doing something along these lines:

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

For the next step they would divide both sides by 2^x but did not cancel the -2 at the end of the equation. This cannot work and was explained to us why in this example.

3+7
----- = 7
3

If we were to follow the idea above then only the 3 would cancel and the answer would be 7 instead of the proper answer being 10/3

We also learned about some other minor bits of information but it's not really important as it was more about some of the graphing calculator functions.

The next scribe on Monday will be Jefferson.