Boyah Forums

2 log₇(x + 8) - log₇(49) = 2

solve itttttttttttttttt

i don't get it

41 5thgrade;

Quote from: Thyme on September 29, 2010, 04:08:47 PM
41 5thgrade;

show me ur proof

how the fuck did you do that?!?! are you magic????

sorry, I never really understand logarithims either. ~_~

u prolly just put it into wolframalpha >:|

Quote from: Nyerp on September 29, 2010, 04:12:35 PM
u prolly just put it into wolframalpha >:|

lol yeah giggle;

does anyone know how to actually do it

i'm 'doing' it


am i doing it right?  we'll see

argh the next question is bullshit and it won't accept my answer

Quote from: Nyerp on September 29, 2010, 04:39:19 PM
argh the next question is bullshit and it won't accept my answer

I miss the good old days of pen and paper. The paper would accept your answer no matter what. :'(

Quote from: Nyerp on September 29, 2010, 04:03:58 PM
2 log₇(x + 8) - log₇(49) = 2

solve itttttttttttttttt

i don't get it

<=>
log₇(x + 8)² - 2 = 2
<=>
log₇(x + 8)²=4
<=>
(x+8)²=7⁴
<=>
x+8=49
<=>
x=41

oh wow i completely ignored the fact that log₇(49) is perfectly solvable goowan

now do this one felty: n_n

log(x + 8) = 1 - log(x - 7)

Quote from: Thyme on September 29, 2010, 04:40:36 PM
I miss the good old days of pen and paper. The paper would accept your answer no matter what. :'(

me too myface;

[spoiler]it was actually 4 months ago in high school where i actually learned things >:[[/spoiler]

Quote from: Nyerp on September 29, 2010, 05:29:43 PM
log(x + 8) = 1 - log(x - 7)

10^(x+8)=1-10^(x-7)
solve as normal

Quote from: TheSequel on September 29, 2010, 07:08:20 PM
10^(x+8)=1-10^(x-7)
solve as normal

doesn't work like that bro

Quote from: Nyerp on September 29, 2010, 07:12:41 PM
doesn't work like that bro

log(y) = log₁₀(y)
lob_b(y)=x

log(x + 8) = 1 - log(x - 7)
10^(x+8)=1-10^(x-7)

Why can't it be rewritten like that?

Quote from: TheSequel on September 29, 2010, 07:22:15 PM
log(y) = log₁₀(y)
lob_b(y)=x

log(x + 8) = 1 - log(x - 7)
10^(x+8)=1-10^(x-7)

Why can't it be rewritten like that?

lob_b(y)=x

to change that to an exponential equation, it's b^x=y, not whatever the hell you just did

you also can't switch to exponential in this case until you combine the logs

so there

holy shit. nyerp is right.

Quote from: Quis sum? on September 29, 2010, 09:33:05 PM
holy shit. nyerp is right.

what

also answer my other question!! plz

Quote from: Nyerp on September 29, 2010, 05:29:43 PM
oh wow i completely ignored the fact that log₇(49) is perfectly solvable goowan

now do this fuck felty: n_n

log(x + 8) = 1 - log(x - 7)

me too myface;

[spoiler]it was actually 4 months ago in high school where i actually learned things >:[[/spoiler]

<=>log(x+8)=log(10)-log(x-7)
<=>log(x+8)=log(10/(x-7))
<=>x+8=10/(x-7)
(x+8)(x-7)=10
x^2+x-56=10
x^2+x-66=0
use quad. formula.

i found the problem i was having

[spoiler]i had read the quadratic formula wrong lol goowan[/spoiler]

thanks giggle;

but stick around cuz i might need more help akudood;

log(x³) = (log(x))²

where did you gooooooooooo

Quote from: Nyerp on September 30, 2010, 07:25:47 PM
log(x³) = (log(x))²

where did you gooooooooooo

3log(x)=(log(x))². Let u=log(x). then 3u=u³. u=0, -sprt(3), sqrt(3). Find the extraneous roots and don't take them into consideration. Find x for each possible u.

just give me the highest zero >.<

ALSO WHAT THE FUCK DOES THE LEWIS STRUCTURE OF A HYPONITRITE ION LOOK LIKE smithicide;

(http://i52.tinypic.com/235tw2.png)

cjlubdoods;

my answer was right already but lol thanks cjlubdoods;

I borked that up. u^2-3u=0. u=0,3

so the zeros are 1 and 1000

yay i got it right

thanks giggle;

I hate math.