Author Topic: onur369's Methods Question Thread :) (Read 30942 times) Tweet Share

luffy · « **Reply #75 on:** March 15, 2011, 11:07:30 pm »

Quote from: Respect on March 15, 2011, 10:29:38 pm

Quote from: onur369 on March 15, 2011, 10:25:01 pm
Quote from: Respect on March 15, 2011, 10:19:12 pm
Quote from: onur369 on March 15, 2011, 10:10:14 pm
Hey guys, got caught up with another question.
I attached it below.

i bet 10 buxs the answer is 2x, lol

No shit sherlock if you check BOB.

nah i checked abbas actually. lol :p

hey idk how to solve it either, wth is it./..

This question is simply trial and error. Just try all the options and see if it satisfies the equation:

a) $f(x) = log_{e}(x)$
$f(\frac{x+y}{2}) = log_{e}(\frac{x+y}{2})$
$\frac{f(x) + f(y)}{2} = \frac{log_{e}(x) + log_{e}(y)}{2} =\frac{log_{e}(xy)}{2}$
$\frac{log_{e}(xy)}{2} \neq log_{e}(\frac{x+y}{2})$
As you can see, this does not satisfy the equation.

However, if you try the same process for option D, the two equations will be equal

brightsky · « **Reply #76 on:** March 15, 2011, 11:12:01 pm »

This looks like some variation of the Cauchy functional equation. By inspection, equations of the form y = ax work because a(x+y)/2 = (ax + ay)/2. However to prove that this is the only possible form is quite hard and I don't know how to do it since x,y = 0 isn't allowed. :/

onur369 · « **Reply #77 on:** March 15, 2011, 11:16:56 pm »

I need help with:
5 + log2 (5x) = log2 (y)
ffs i have problems converting 5 into a log2 form

brightsky · « **Reply #78 on:** March 15, 2011, 11:22:49 pm »

5log2(2) + log2(5x) = log2(y)
log2(2^5) + log2(5x) = log2(y)
log2(2^5*5x) = log2(y)
y = 2^5*5x = 140x

onur369 · « **Reply #79 on:** March 15, 2011, 11:26:16 pm »

Thanks mate, definitely a 99.95 in 2013 from you, extremely smart. Keep it up ! =)

nacho · « **Reply #80 on:** March 16, 2011, 02:12:31 pm »

Quote from: onur369 on March 15, 2011, 11:16:56 pm

I need help with:
5 + log2 (5x) = log2 (y)
ffs i have problems converting 5 into a log2 form

I'm not bashing on you, but i suggest that you have another look at how the log works, it is quite crucial for you to know these first-level basics if you want to go on and do harder problems.

Mao · « **Reply #81 on:** March 17, 2011, 11:31:21 pm »

Moderator Action: off-topic posts have been deleted, original advice by nacho is preserved. Everyone whose post(s) have been deleted, consider this a serious reminder.

C'mon guys, this is the maths board for crying out loud.

Please keep these confrontations in pm, you guys (and ladies) are mature enough to be able to sort these things out without making a racket. I'm sure a small misunderstanding such as this can be settled by a simple message exchange. There's no need to publicize it and give it the 'wolf pack' feeling.

If you would like to voice without a direct confrontation, or feel a pm is too direct, please do not hesitate to contact any of the mods and we will help to mediate.

Back on topic please. =]

Respect · « **Reply #82 on:** March 17, 2011, 11:40:50 pm »

Quote from: Mao on March 17, 2011, 11:31:21 pm

Moderator Action: off-topic posts have been deleted, original advice by nacho is preserved. Everyone whose post(s) have been deleted, consider this a serious reminder.

C'mon guys, this is the maths board for crying out loud.

Please keep these confrontations in pm, you guys (and ladies) are mature enough to be able to sort these things out without making a racket. I'm sure a small misunderstanding such as this can be settled by a simple message exchange. There's no need to publicize it and give it the 'wolf pack' feeling.

If you would like to voice without a direct confrontation, or feel a pm is too direct, please do not hesitate to contact any of the mods and we will help to mediate.

relax, mao... we're just messing, nacho is our freind, dont u like to tease ur freinds sometimes

rite nacho?

Mao · « **Reply #83 on:** March 17, 2011, 11:57:01 pm »

Quote from: Respect on March 17, 2011, 11:40:50 pm

Quote from: Mao on March 17, 2011, 11:31:21 pm
Moderator Action: off-topic posts have been deleted, original advice by nacho is preserved. Everyone whose post(s) have been deleted, consider this a serious reminder.

C'mon guys, this is the maths board for crying out loud.

Please keep these confrontations in pm, you guys (and ladies) are mature enough to be able to sort these things out without making a racket. I'm sure a small misunderstanding such as this can be settled by a simple message exchange. There's no need to publicize it and give it the 'wolf pack' feeling.

If you would like to voice without a direct confrontation, or feel a pm is too direct, please do not hesitate to contact any of the mods and we will help to mediate.

relax, mao... we're just messing, nacho is our freind, dont u like to tease ur freinds sometimes

rite nacho?

As I said, please keep it on topic.

kefoo · « **Reply #84 on:** March 19, 2011, 01:50:39 pm »

Hey guys, need abit of help here. I know this q may sound simple, but dunno how to do it

edit: didnt see it intersected at (0,0)
another question though:

Prove that if logr(P) = q and logq(r) = p then logq(p) = pq

pi · « **Reply #85 on:** March 19, 2011, 02:41:47 pm »

$Prove \ that \ if \ \log_{r}(p) = q \ and \ log_{q}(r) = p \ then \ log_{q}(p) = pq.$

$\log_{r}(p) = q \ \cdots \ (1)$
$\log_{q}(r) = p \ \cdots \ (2)$
$Using \ (1) \ r^q = p \ \therefore \ r = p^{1/q}$
$Using \ (2) \ r=q^p$
$\therefore \ q^p = p^{1/q}$
$\therefore \ q^{pq} = p$
$\therefore \ \log_{q}(p)=pq$

There's probably an easier way, but meh, this one works

kefoo · « **Reply #86 on:** March 19, 2011, 03:12:29 pm »

Quote from: Rohitpi on March 19, 2011, 02:41:47 pm

$Prove \ that \ if \ \log_{r}(p) = q \ and \ log_{q}(r) = p \ then \ log_{q}(p) = pq.$

$\log_{r}(p) = q \ \cdots \ (1)$
$\log_{q}(r) = p \ \cdots \ (2)$
$Using \ (1) \ r^q = p \ \therefore \ r = p^{1/q}$
$Using \ (2) \ r=q^p$
$\therefore \ q^p = p^{1/q}$
$\therefore \ q^{pq} = p$
$\therefore \ \log_{q}(p)=pq$

There's probably an easier way, but meh, this one works

nice! thanks alot
just one more if thats ok
y = ae^x + b with points (1,14) and (0,0)
i did simultaneous eqns but i get the wrong answer =.=

pi · « **Reply #87 on:** March 19, 2011, 03:20:31 pm »

$y = ae^x + b \ with \ points \ (1,14) \ and \ (0,0)$

$Using \ (0,0) \therefore \ 0 = a + b \ \therefore \ a=-b \ \cdots \ (1)$
$Using \ (1,14) \therefore \ 14 = ae +b \ \cdots \ (2)$
$Sub \ (1) \ into \ (2)$
$14 = -be +b$
$14 = b(1-e)$
$b = \frac{14}{1-e}$
$\therefore \ a=-b = -\frac{14}{1-e}$

Hopefully thats right

kefoo · « **Reply #88 on:** March 20, 2011, 11:36:53 am »

Quote from: Rohitpi on March 19, 2011, 03:20:31 pm

$y = ae^x + b \ with \ points \ (1,14) \ and \ (0,0)$

$Using \ (0,0) \therefore \ 0 = a + b \ \therefore \ a=-b \ \cdots \ (1)$
$Using \ (1,14) \therefore \ 14 = ae +b \ \cdots \ (2)$
$Sub \ (1) \ into \ (2)$
$14 = -be +b$
$14 = b(1-e)$
$b = \frac{14}{1-e}$
$\therefore \ a=-b = -\frac{14}{1-e}$

Hopefully thats right

ohh thats how you do it thanks.
weekend is probably getting to me =.=

kefoo · « **Reply #89 on:** March 22, 2011, 06:38:10 pm »

stuck on another question...

if f(x) = 1 - e^(-x)
find the inverse.

i swap the x and y as usual, but dunno how to get the log and stuff..

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