Kadushy Posts
Kadushy
Douchebag
lrn2gramcrackers
Kadushy
Douchebag
Lol.
Kadushy
Douchebag
Lol.
Kadushy
Douchebag
k, sorry.
Kadushy
Douchebag
Kadushy
Douchebag
I slept 14 hours today because I didn't want to deal with my hangover.
- November 23, 2012
- November 23, 2012
Kadushy
Douchebag
Edit: Shit, I've been inb4'd
Kadushy
Douchebag
all i got was a hangover.
Kadushy
Douchebag
:)
...
...
Kadushy
Douchebag
k das kewl
Kadushy
Douchebag
awws hit
Kadushy
Douchebag
go away
Kadushy
Douchebag
Omg what a great deal. I'll definitely buy one when I save up enough.
Kadushy
Douchebag
callintz_klacid wrote...
Spoiler:
You did great, only a bit off. It wants you to prove from
=Log[1/3 (a+b)]
and change it into
=1/2[log a +log b]
Here's how i think it should be:
Spoiler:
Sorry for the rotation...
You're awesome. Thanks! lol
Kadushy
Douchebag
I haven't slept yet... and it's 7am.. that's what I hate about today.. and it's Thanksgivings/Black Friday. I guess I'll stay up since I needed to fix my sleeping schedule yet again.
Kadushy
Douchebag
Sounds fun..
Kadushy
Douchebag
I received a few things in the past 7 days... which are...
-Gundam Exia 1/144 scale ($18~)
-Velvet Kiss Volume 2 with a Fakku sticker ($13~)
-Asus VH236H 23" 1920 x 1080 Monitor ($182, and I'm still waiting on my $20 rebate... I hope I did it right.. dat UPC code)
It's a pretty good week... though not really since I failed my test D:
-Gundam Exia 1/144 scale ($18~)
-Velvet Kiss Volume 2 with a Fakku sticker ($13~)
-Asus VH236H 23" 1920 x 1080 Monitor ($182, and I'm still waiting on my $20 rebate... I hope I did it right.. dat UPC code)
Spoiler:
It's a pretty good week... though not really since I failed my test D:
Kadushy
Douchebag
If a² + b² = 7ab, show that Log [1/3 (a+b)] = 1/2 (Log a + Log b)...
So... (a+b)² = a² + 2ab + b² then group them into "a² + b² +2ab" and replace "a² +b²" with 7ab since the "if = 7ab" to make 7ab + 2ab... which equals 9ab...
Then... (a+b)² = 9ab... take the square root of both sides to get (a+b) = 3√ab
After that use the log properties.. then uh...
Log (a+b) = Log 3 + Log a½ + Log b½... then subtract Log 3 on both sides to get Log (a+b) - Log 3 = Log a½ + Log b½...
By using the log properties "Logь (x/y) = Logьx - Logьy" and "Logь(xy) = Logьx + Logьy" and "Logьx^r = r Logьx" I'll end up with...
Log (a+b)/3 = ½ (Log a + Log b)
[1/3 (a+b)] = (a+b)/3....
So... uh... is this it? Or do I have to solve it? It says show... and I think I showed it but I'm not sure...
So... (a+b)² = a² + 2ab + b² then group them into "a² + b² +2ab" and replace "a² +b²" with 7ab since the "if = 7ab" to make 7ab + 2ab... which equals 9ab...
Then... (a+b)² = 9ab... take the square root of both sides to get (a+b) = 3√ab
After that use the log properties.. then uh...
Log (a+b) = Log 3 + Log a½ + Log b½... then subtract Log 3 on both sides to get Log (a+b) - Log 3 = Log a½ + Log b½...
By using the log properties "Logь (x/y) = Logьx - Logьy" and "Logь(xy) = Logьx + Logьy" and "Logьx^r = r Logьx" I'll end up with...
Log (a+b)/3 = ½ (Log a + Log b)
[1/3 (a+b)] = (a+b)/3....
So... uh... is this it? Or do I have to solve it? It says show... and I think I showed it but I'm not sure...
Kadushy
Douchebag
Here's me begging someone out there who can help me solve this since gah dayuuuum I suck at math and especially on logs. Pretty please?
inb4 do your own homework, lock
Edit: I just googled and found this... and lots more... http://www.algebra.com/algebra/homework/logarithm/logarithm.faq.question.313421.html
but dayum I need to know this stuff since ...finals..
^Still confused as shit. don't even know how they get that 2ab on the right side.
inb4 do your own homework, lock
Edit: I just googled and found this... and lots more... http://www.algebra.com/algebra/homework/logarithm/logarithm.faq.question.313421.html
but dayum I need to know this stuff since ...finals..
^Still confused as shit. don't even know how they get that 2ab on the right side.
