anonymous
  • anonymous
LOG(b2) 3rdroot/2 (typed version after the jump)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\log _{2} \sqrt[3]{2}\]
anonymous
  • anonymous
|dw:1352348272404:dw|
anonymous
  • anonymous
What can you do next?

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anonymous
  • anonymous
get rid of the base 2 and the two in the parenthesis?
anonymous
  • anonymous
You have a power in a log.
anonymous
  • anonymous
|dw:1352348442988:dw|
anonymous
  • anonymous
hmm, ok. my teacher added an =X to it first, and then took off the log and made it 2^x
anonymous
  • anonymous
Lot harder...
anonymous
  • anonymous
|dw:1352348582297:dw|
anonymous
  • anonymous
|dw:1352348653250:dw| and then...
anonymous
  • anonymous
Okay that works too...
anonymous
  • anonymous
My way is a lot easier I think though.
anonymous
  • anonymous
|dw:1352348698633:dw|
anonymous
  • anonymous
cool, do you think you could come up with a new example that would work?
anonymous
  • anonymous
Sure.
anonymous
  • anonymous
awesome, that way I can get a fresh go at it
anonymous
  • anonymous
|dw:1352348747232:dw|
anonymous
  • anonymous
Try that.
anonymous
  • anonymous
That's a pi/4 btw.
anonymous
  • anonymous
hm, ok...
anonymous
  • anonymous
|dw:1352348848985:dw| I did it her way, just cuz I've been doing that other one like that...
anonymous
  • anonymous
Okay fine. :p . Hmm...
anonymous
  • anonymous
that anything close to being correct? haha
anonymous
  • anonymous
It is correct.
anonymous
  • anonymous
|dw:1352349005378:dw|
anonymous
  • anonymous
That's a 2*pi btwe.
anonymous
  • anonymous
btw*
anonymous
  • anonymous
ok, I haven't done any of these with pi, so do I just treat it like a number, like 3.1416 or something?
anonymous
  • anonymous
yeah :p .
anonymous
  • anonymous
I prefer exact answers but sure.
anonymous
  • anonymous
I think you get the point though so I will stop. Good job :) .
anonymous
  • anonymous
haha, this whole concept of log is just really new and confusing to me
anonymous
  • anonymous
You are doing fine :) .
anonymous
  • anonymous
I'm just doing the steps as best I can, but I have no clue what it all means. hoping the understanding will come later lol. lemme see what I can do with that one
anonymous
  • anonymous
|dw:1352349332360:dw| pretty sure I messed that up
anonymous
  • anonymous
Yeah you did :P .
anonymous
  • anonymous
haha
anonymous
  • anonymous
|dw:1352349512686:dw|
anonymous
  • anonymous
That was 2 * pi not 2*x*pi .
anonymous
  • anonymous
Anyways, I am getting ahead of myself. You will learn all this soon.
anonymous
  • anonymous
probably so haha. see if you can help me with one like this...
anonymous
  • anonymous
|dw:1352349693033:dw|
anonymous
  • anonymous
|dw:1352349742608:dw|
anonymous
  • anonymous
I should mention: |dw:1352349777605:dw|
anonymous
  • anonymous
log base a to the ath power equals 1?
anonymous
  • anonymous
Log base a of a is 1.
anonymous
  • anonymous
how did you get 1/25 to 5^-5
anonymous
  • anonymous
Reciprocal values.
anonymous
  • anonymous
Sorry I should have said 5^-2.
anonymous
  • anonymous
|dw:1352349972862:dw|
anonymous
  • anonymous
ah
anonymous
  • anonymous
|dw:1352349999623:dw|
anonymous
  • anonymous
ok I see
anonymous
  • anonymous
|dw:1352350166704:dw| so, on parts like this, are you actually "cancelling" the fives? thats how I've understood/been working them.
anonymous
  • anonymous
Well Like I said: |dw:1352350268114:dw|
anonymous
  • anonymous
Proof:|dw:1352350284036:dw|
anonymous
  • anonymous
haha, I feel like I'm so close to getting that.
anonymous
  • anonymous
Okay we know that logs are the inverse of exponential function right?
anonymous
  • anonymous
|dw:1352350446240:dw|
anonymous
  • anonymous
instead of like 1,2,3,4,5 they can represent a different measure of counting, is how I understand them. yes?
anonymous
  • anonymous
Not really...
anonymous
  • anonymous
Logarithms are just the inverse functions of exponential functions.
anonymous
  • anonymous
I'm not quiet sure I follow. I'm close though ha
anonymous
  • anonymous
Okay. |dw:1352350715897:dw|
anonymous
  • anonymous
They are simply inverses.
anonymous
  • anonymous
so like 5^2 is 25, the inverse would be 25 something =5?
anonymous
  • anonymous
the "something" is the part that I'm missing lol
anonymous
  • anonymous
|dw:1352350885891:dw|
anonymous
  • anonymous
awesome, now if I can just remember that. :-)
anonymous
  • anonymous
You will :) .
anonymous
  • anonymous
I guess I'm just trying to get my brain to read "log base 5 to the 25 equals 2" in a way that tells me exactly what i'm trying to do to it
anonymous
  • anonymous
I don't know why thats so difficult for me. it's very annoying
anonymous
  • anonymous
I found logs hard the first time too :) .It's fine. You will get the hang of it.
anonymous
  • anonymous
my teacher goes really fast, so that doesn't help. I mostly just copy everything down, and don't have a chance to think about it
anonymous
  • anonymous
Same Same... I find doing practice problems helps the most.
anonymous
  • anonymous
yea. just trying to take my time and get it to develop into a solid understanding.
anonymous
  • anonymous
|dw:1352351467721:dw| ok, I'm going to try this one, I think I might have it
anonymous
  • anonymous
|dw:1352351563158:dw|
anonymous
  • anonymous
dang. thats not quite it... forgot I had the answer sheet
anonymous
  • anonymous
That right! :O .
anonymous
  • anonymous
Oh, youre right, I just didn't follow it through to x=32 like she did
anonymous
  • anonymous
|dw:1352352361556:dw| it's starting to make a little sense finally
anonymous
  • anonymous
Yep correct :) .
anonymous
  • anonymous
|dw:1352352553761:dw| do I need to multiply the fraction out?
anonymous
  • anonymous
No.
anonymous
  • anonymous
|dw:1352352659947:dw| I'm stuck on the 3rd root part
anonymous
  • anonymous
|dw:1352352937727:dw| isn't that how it works?
anonymous
  • anonymous
hm, no I did something wrong.
anonymous
  • anonymous
|dw:1352353487129:dw| well, I think that will be all I can do tonight. Thanks so much, you are awesome and I really appreciate it.

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