Wednesday, November 14, 2007

Logarithms and Exponents

Hey guys, it's Luis posting. Today we start off talking about narrowing our votes from 16 down to 7, to belong some of our scribe posts into the hall of fame...and we decided that 2 points for the outsiders, mentors or teachers. And then Mr. K had given us some problems to sketch their graphs... here are they are:








We should be able to picture what y= e^x looks like.. and WE can easily figured out two points (0,1) and (1,e), e which is a little bit of less than three. In y=e^x-2, we just have to shift our graph two units down. our asymptote is y=-2, our graph should not cross over that line.
In number two problem, draw first an imaginary line of y=e^x, then -e = we just have to flip it from up to down but over the x-axis because e is negative, -x= we just flip it from right to left. and our asymptote is y=0.

In the last problem, we have to start ignoring the absolute value, then graph the e^x-1...then the last thing we should do is to figure out what is the absolute value. (that was easy! huh.) Our asymptote this time is not equal to -1, it should be 1 because of the absolute value.

Anyways, our prof then discussed the properties of Exponential and Natural Log of Functions by comparing what y=e^x and y=ln(x) looks like...visit our number one graphing online tool,







Whapak!, yes! we're on the right track, y=ln(x) is the inverse function of y=e^x. Haha. I never thought of that, do you? Anyways, we then easily understand what is the difference of their domain and range, we just have to switch them.

Next stop, sketching the graph of... (it's fun though, right?)


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One of our classmate, answered this question easily...BUT...but Mr. K. is looking for the two points, (1,0) and (e,1) which (-1,0) and (-e,1). And our asymptote is x=0 because it is ln. Her answer was brilliant though.
Then Mr. K had given us a problem, which terrified us (not really, hehe)...but confused us a lot and asked him lots of questions...

But then explained us that the exponent on the base e that gives us e^x... will give us x. We should really recognize that ln of e^x is equal to x. Then apply the power law...Even though we used the common logarithm (right side) their answers are both the same.


Before our period ends, Mr. K. had introduced us to our new topic, Exponential Modeling. Bare with me guys, but I didn't get most of it...but I'm pretty sure he will talk about this again tomorrow. Ask him lots of questions...OK? And don't forget guys, Logarithms are Exponents.

That's all about we learn this morning...The next Scribe is Jordan.

1 comment:

Julia.h.s.student. said...

sorry i got kinda lost after you distributed the (ln2) throughout the (x+1).
x=(1-ln2)+ln2
how did you get from that to
x(1-in2)=ln2?
oh.nevermind. i got it thanks.