Vastor

thnx for both of ya, actually, ajb's short explaination already confirmed my thought about the phytagoras "shape" on the graph itself. but the explaination from Appolinaria is wonderful... about other equation/formula, I may derived myself to find the intuitive behind it, though so will visit this post again if had any question...

hey guys, from the (math)textbook, The distance between the points [math]P(x_1, y_1)[/math] and point [math]Q(x_2, y_2)[/math] is [math] \sqrt{(x_2 - x_1)^2 + (y_2- y_1)^2}[/math] what does the intuition behind the statement? it look like much Phytaghoras theorem, does anything related here? another thing is the intercept form [math] \frac{x}{a} + \frac{y}{b} = 1 [/math], where a = x-intercept, and b = y-intercept. anyone can give how does this formula formed from? or atleast any intuitive behind this equation? hope anyonce can help, btw, if anyone got good website/tutorial about coordinate geometry, please send me links.. thnx

don't you think there error in your calculation? how do you get x - (3-1) = 3/x ?

hey guys, ignore my post before and never use your pc calculator to calculate log xD anyway, I got another sort of hard question: Solve the equation [math] 4^x - 7*2^x - 8 = 0[/math] [math] 4^x - 7*2^x - 8 = 0 [/math] [math] 4^x - 7*2^x = 8 [/math] [math] 2^x(2^x - 7) = 2^3 [/math] [math] 2^x - 7 = \frac{2^3}{2^x} [/math] using trial and error, you can get x = 3 but how to calculate it?

then, It should be, the harder you try NOT to sleep, the faster you sleep.

how can this happened?

you spell it right now, no more problem then but, today, after doing some exercise, I'm still miss-read some constant.... again! should I start drink tea instead of coffee now?! oh btw, in exam, there no answer on the back, so I can't check my mistake... :lol: :lol:

haha, that(Sod's Law) should explain everything doesn't it? thnx actually, I'm not taking this thread seriously, but well, I'm not really stressed for the incident (probably because I love the subject so much). Eventhough those event is kind of unlucky, at least I'm learning something from it (completely understanding log for my syllabus, except for the base part, though), and that's matter the most. But it's certainly unlucky to do the mistake in exam or test. When thinking how badluck I am(shorty, poor family, not that good-looking), I always comfort myself with thought about how much person which has worse luck than mine, is this kind of hypocrite?!

symptom - more pimple pop on my face. - my computer BSOD this day. - read [math]3^{t-2}[/math] as [math]3^{t-3}[/math] - solve [math]9x - 8 = 0[/math] as [math]x = \frac{9}{8}[/math] - calculate [math](k^2)(k^{5+t})[/math] as [math] k^{2(5+t)}[/math] - wrong almost every problem that I solved by mistake from above(except for pimple). - losing my calculator. - god is not exist. - not having my coffee today. - having "eh, I know how to answer this" moment while writing long post to ask the expert here. I have two question, did I'm still sane? please said to me the above symptom is just bad luck, right?! should I stop try to be funny?

oh, my bad, never found it before even already checked profile setting before

hey guys, I'm still having this problem, this is a bug or something? anything I can do? should I make new account not based on facebook so that I can change the avatar pic?

edit : Ignore this post, should have read question better hey guys, I don't know if the problem is in the answer provided in the book but the exercise said Solve the following equation [math] 9log_3 x = 7 [/math] [math] 9log_3 x = 7 [/math] [math] log_3 x = \frac{7}{9} [/math] [math] 3^{\frac{7}{9}} = x [/math] answer in the book [math] x = \sqrt{7}[/math] ooo wait, does it legitimate to use graphing calculator like this to find the answer? lost my calculator somewhere

forget this, this one is clue on how to solve the question's problem... I understand that, when first seeing log, I can't really see the pattern on how it work, and can't even understand why [math]A^B = C [/math] become [math] log_A C = B[/math] until now, the only thing not really clear to me is what does A represent (which why give me so much difficulty on converting from indice to log or otherwise), yes it is "base" but what is it actually, does it refer to the same "base" of "number type"(sorry, not really know many definition on math here), I mean does it refer to base-ten / base-two number? if so, why does [math]log_1 12^1 = 12[/math] does not 12 is number base-10? why it is 1 in log? for the part that I understand on log:- - they behave like indices, except they are being as the "main number". example:- [math]10^5 = 100,000[/math] become [math]log 100,000 = 5[/math] the 5 became the "main number", and the "original main number" become inner-side of the log. - we can change the number base(and the "original main number") as we like without changing its "main number" values. examples:- [math]1 = log 10 = log_5 5 = log_6 6 = 1[/math] - there some pattern on how there some relation between the "base" and "main number" and "inner-side number"(all number, nuff said)in log [math]log_9 \frac{x}{y}[/math] [math]=\frac{log_3 \frac{x}{y}}{log_3 9}[/math] [math]=\frac{log_3 \frac{x}{y}}{2}[/math] [math]=\frac{log_3 x - log_3 y}{2}[/math] [math]=\frac{1}{2}(log_3 x - log_3 y)[/math] [math]=log_3 \sqrt{x} - log_3 \sqrt{y}[/math] can you say something about this?(wikipedia just not enough) edit: should proof-read first

ok, I got it, thnx, so the clue here is to put the variable x in one place only(aka not seperated between two number by the plus symbol)?!

anyone?

hey guys, I'm login from my fb, when I change my avatar pic to other pic, on the next day it change back to my fb account pic, how can I fix this?

ouch, I don't know what the cause, my brain hurt so hard when dealing with logs that involve adding like this question, indices is much better alternative [math]3^{x-1} + 3^x = 12 [/math](originals of above question) using factorisation and other stuff without involving log can be solved easily and get the answer of x = 2. but can log solve above problem also? I try but I had problem when I'm trying to convert the indices to a log because of the add/minus symbol. it give me so much headache on how to create a symbol for the add/minus (like multiply and divide they turn into plus/minus in log) if then, what plus/minus turn out to be? even using ajb recommendation, I still can't convert it in which I'm sure I'm do it right if [math] x = \log_{a}(x^{a}) = a^{\log_{a}(x)} [/math] then [math] 3^{x-1} + 3^x = \log_{1}{12^1}[/math] i'm doing it right? if then, how I convert on the left part to log? arrrghh

hey guys, well, as we know, adding among log like this is easy [math]\log_c A + \log_c B = \log_c AB[/math] but how does to "add" something like this? [math]\log_3 12-3^x = x - 1[/math] and basically, how you calculate add in an equation that has indices and logarithms? I mean if [math]A * A = A^{1+1}[/math] but how about [math]A + A = A^? = 2A[/math]

simple answer that explain all, thnx hope I'm doing good now:- [math]C(x-a)^2[/math] [math]Cx^2 - 2aCx + Ca^2 = 0 [/math] compare [math]p^2x^2 - 3qx +4 = 0[/math] [math]C = p^2[/math] so, [math]p^2a^2 = 4[/math] [math]a^2 = \frac{4}{p^2}[/math] [math]a = \frac{\pm2}{p}[/math] then, [math]-2aC = -q[/math] [math]2(\frac{\pm2}{p})(p^2) = 3q[/math] p, q > 0 so, [math]2(\frac{2}{p})(p^2) = 3q[/math] [math]4p = 3q[/math] [math]\frac{4}{3} = \frac{q}{p}[/math] another problem! given that [math]a[/math] and [math]b[/math] are the roots of the equation [math]3x^2 + 7x - 6 = 0[/math] where [math]a > 0[/math] and [math]b < 0[/math]. Form a quadratic equation which has the roots [math]a + 3[/math] and [math]b-2[/math]. my progression:- [math]3x^2 + 7x - 6 = 0[/math] [math]x^2 +\frac{7}{3}x - 2 = 0[/math] compare [math]x^2 - (a+b)x + ab = 0[/math] so, [math]a+b = -\frac{7}{3}[/math] & [math]ab = -2 [/math] for, [math]ab = -2 [/math] [math]a = -\frac{2}{b} [/math] and then, for [math]a+b = -\frac{7}{3}[/math] [math] -\frac{2}{b}+b = -\frac{7}{3}[/math] [math] b = \frac{2}{b} - \frac{7}{3} [/math] [math] b = \frac{6 - 7b}{3b} [/math] [math]3b^2 + 7b - 6 = 0[/math] [math](3b - 2)(b + 3) = 0[/math] [math]b = \frac{2}{3}[/math] and [math]b = -3[/math] because [math]b < 0[/math], so [math]b = -3[/math] only. after that, for [math]a = -\frac{2}{b} [/math] [math]a = \frac{2}{3} [/math] then, [math]x^2 - (a+3 + b-2)x + (a+3)(b-2) = 0[/math] [math]x^2 -(-\frac{7}{3} + 3-2)x + (ab+3b - 2a-6) = 0[/math] [math]x^2 + \frac{4}{3}x +(-8 + 3b - 2a) = 0[/math] [math]x^2 + \frac{4}{3}x +(-8 + 3(-3) - 2(\frac{2}{3})) = 0[/math] [math]x^2 + \frac{4}{3}x +(-17 - (\frac{4}{3})) = 0[/math] [math]x^2 + \frac{4}{3}x - \frac{55}{3} = 0[/math] [math]3x^2 + 4x - 55 = 0[/math] and the answer is [math]3x^2 + 4x - 55 = 0[/math] yes, it's the same but I just can't figured out how does the [math]b = \frac{2}{3} = a[/math] or it's just a coincidence?!

not really, just typo transferring from the paper that I used to calculate, and not missing any [math]x^2[/math] there at [math] \frac{\pm 2}{p} = a[/math] why not the p has plusminus value when squareroot it? EDIT : Ignore this post

The equation [math](px)^2 + 3qx + 4 = 0 [/math] has two equal roots. Given that [math]p > 0[/math] and [math]q > 0[/math], find [math]p : q[/math]. my progression:- [math]p^2x^2 + 3qx + 4 = 0 [/math] where [math]x = a[/math] (two equal roots) [math](x-a)^2 = 0[/math] [math]x^2 - 2ax + a^2 = 0[/math] compare [math]p^2x^2 + 3qx + 4 = 0 [/math] [math]a^2 = 4[/math] [math]a = \pm2[/math] if [math]a = 2[/math] [math]-2a = 3q[/math] (compare) [math]-4 = 3q[/math] [math]q = \frac{-4}{3}[/math] [math]\pm[/math] effect ( if a = -2) [math]q = \pm\frac{4}{3}[/math] but, q > 0, so no minus [math]q = \frac{4}{3}[/math] [math]p^2x^2 + 4x + 4 = 0[/math] discriminant (two equal roots) [math]4^2 - 4(p^2)(4) = 0[/math] [math]16- 16p^2 = 0[/math] [math] p = \pm 1[/math] but, p > 0, so no minus [math] p = 1[/math] answer on the book [math]p : q = 3 : 4[/math] anyone can help? edit: p^2 + 3qx + 4 = 0, sorry about the typo :/

thnx for clarify it! heh, I knew it that he is no different with teacher that teach math by making student memorize each formula and step rather than teach on how to make student understand what they are talking about... talking about current science education, don't know who to be blame, but maybe filling exam with this kind of question can change the style of teaching(and learning) :- rather than putting a silly quadratic equation and solve for its x. and with current education also, it mislead some people(not remember who say this on lowyat forum) *wtf?!* thnx for recommendation, I think self-teaching is not problem much compared to having teacher that can't teach. can ask here if not understand any also for biology and phyisc, anyone have recommendation?(book, e-book, tutorial etc...)

oh, so that's what you mean by "real", ha3, well, I thought non-real = undefined(x = 1, etc...), but talking about my "non-existant" knowledge of imaginary number, I'm just self- teaching about it, never taught at school yet... think I just leave the case from me...

SPM(O-level equivalent) has 9 subject, archieving 8A+ and above guaranteed a place of scholarship by government where A+ > 89% for every subject. k, I'm trying to archieve good result so that I can get scholarship to continue my studies for A-level(which considered to be expensive here) rather than going for STPM(A-level equivalent) with less than 10% cost, but "General Paper" is compulsory in exam, I hate being forced to learn anymore, enough at SPM already(5 subject compulsory) but on my current condition now, time is very demanding, (should not waste my time anymore T_T) so, yeah, need learning source other than Khan Academy , that effective to learn chemistry and physic. or maybe there some tip on how to study and archieved mastery in a fast way? what I know, in SPM, the math is proficiency based (rather than logical skill based), I mean, the time is too short, rather than a question about real-life(word problem, etc...) if you not remember formula = you're doomed!, bad news for me, talking about my bad memories brains... x( in physic, logical thinking demand more though so there are memory based question,(talking about remembering definition, lol, unlikely for anyone remembering definition can proceed anything to understand the definition itself) and I got some sort of bad teacher in chemistry, rather than teaching us on how to form the equation of any reaction on some sort, he told us to remembered it, lol, (I MEAN EACH EQUATION ) and btw, I'm self-studying C++ currently, should I leave to concentrate my study on these subject? I just spent 2 hours daily on it so that I'm not forgot it at all, I hear some say that it's good for mathematician / physicist to learn at least one programming language... what do you think?