7 ms·

Trouble with logarithms and partial fractions

2 points by pencil 16y ago

I'am in my 30's who's willing to relearn math from ground up and i really felt HN is the right place to get my problem solved so i've decided to post a couple of mathematical problems which i'am having trouble with. if log(a-b/4) = log sqrta+log sqrtb,show that (a+b)^2 = 20ab if x^2+y^2=8xy show that 2log(x+y) = log5+log2+logx+logy decompose the following into partial fractions 1) 3x-1/[(x+2)(1-x+x^2)] 2) 1/(x^3+1) 3) 2x^2-14x+8/[(x^2+3x-2)(x-3)]

pencil 16y ago
i couldn't find any book that clearly explains logarithms and partial fractions nor khan acadamy was able to find a solution to this.
- RiderOfGiraffes 16y ago
  Did you try this one? http://www.khanacademy.org/video/introduction-to-logarithms http://www.khanacademy.org/video/introduction-to-logarithms Or this one: http://www.khanacademy.org/video/partial-fraction-expansion-1 http://www.khanacademy.org/video/partial-fraction-expansion-...
  - pencil 16y ago
    ya i've tried them
ghoul2 16y ago
the second one is easy: it is given that X^2+Y^2=8XY, thus, X^2+Y^2+2XY=10XY (i.e, added 2XY to both sides) thus, (X+Y)^2 = 10XY. Taking log of both sides: 2 log (X+Y) = log 10 + log X + log Y = log 5 + log 2 + log X + log Y
ghoul2 16y ago
heh, come to think of it, so is the first one: it is given that log((a-b)/4) = log sqrta + log sqrtb, thus (a-b)/4 = sqrt(ab), thus, ((a-b)/4)^2 = ab, thus, (a-b)^2 = 16ab, thus a^2 - 2ab + b^2 = 16ab. Adding 4ab to both sides, a^2 + 2ab + b^2 = 20 ab, giving (a+b)^2 = 20ab.
- pencil 16y ago
  Thank you so much.what about the rest??any idea??
  - ghoul2 16y ago
    the others are even easier than these. I think if you are really trying to get back into math you should not lose patience so easily and spend a bit more time trying, and don't ask for help even if it takes many days to get these questions - otherwise it will never click for you again (my own personal experience). I also hope I have not given you full solutions to homework problems you were supposed to solve by yourself. The wikipedia page on partial fractions gives you good directions on how to solve it.
RiderOfGiraffes 16y ago
What have you tried? What do you know? Rearrange 2.log(a)=3.log(b)+5 to find b in terms of a. Answering that will give us some idea of your current status the better to answer your questions.
- pencil 16y ago
  thank you so much for your response,actually rearragning the terms to find 'b' gives me 3logb = 2loga - 5 {upon subtracting 5 from both the sides}. this is where i need help.please look into my partial fractions problems as well if you got the time.
  - RiderOfGiraffes 16y ago
    But you still don't have b as the subject of the equation. You need to get to "b=..." If you don't know how to do that, then at least tell us what you've tried. Can you make x the subject of this equation: log(x)+3=2t ? In other words, if the equation log(x)+3=2t is always satisfied, what does x equal? And I will look at the partial fractions, but I'd like to concentrate on this first. Actually, what is your definition of a logarithm? The last time we had a conversation I asked you some questions to assess your level of knowledge and you never answered them: http://news.ycombinator.com/item?id=1107996 http://news.ycombinator.com/item?id=1107996 It seems to me looking over other times you've asked questions that you claim you want to get into math, and you probably think you're doing lots of stuff, but in reality I've seen no evidence of you actually trying things. Gaining skills is a matter of spending the time. In math in particular, if you haven't spent ages getting loads of wrong answers then you won't have any real intuition as to what to do. You need loads of wrong answers, from which you can start to identify "good wrong answers". From those good wrong answers you can build an intuition as to what might work, and you can start to find excellent wrong answers. Then the moment comes when you see how to convert an excellent wrong answer into a right answer. With practice this becomes second nature, and after a time you wonder why you ever found it difficult. But until you've put in the hours of practice, it doesn't happen. People think doing math is a matter of somehow finding the right process and then following it. "Tell me what to do" is the perpetual cry. You need to try stuff. Try re-arranging the equations repeatedly into different forms until, maybe, one looks familiar. Don't just stare at stuff hoping inspiration will strike, and don't think there's a magic formula for finding the right next step. That's not how it works. So tell me, how many different ways can you rearrange the equation: log(x)+3=2t ?
RiderOfGiraffes 16y ago
1) 3x-1/[(x+2)(1-x+x^2)] 2) 1/(x^3+1) 3) 2x^2-14x+8/[(x^2+3x-2)(x-3)] In 1 and 3 are there supposed to be extra brackets? It's really hard to tell exactly what these are given that you've used no formatting. Secondly, why are you trying to solve these? Where did they come from? Solving the first one, assuming the numerator is supposed to be (3x-1), then we proceed as follows. When we add fractions like a/b + c/d the result is (ad+bc)/bd. In these questions we assume we have the result of such an addition and try to compute all the terms. Here we have a denominator of (x+2)(1-x+x^2) so we assume b=x+2 and d=1-x+x^2. that means we have the numerator formula is: (3x-1) = a(1-x+x^2) + c(x+2) In this case a and c are potentially polynomials (in fact by looking at the degrees c must be a polynomial of degree 1 larger than a). Now just try setting a and c to small polynomials, c being 1 degree larger, and compute what you get on the right hand side. Does it look like the left hand side? Probably not. What do you need to change? How can you make it better. Jiggle it around a bit and see what happens. Tell me, what if a=(x+1) and c=(x^2-2) : what does the RHS evaluate to? Why is it wrong? How can you change a and c to make it better. With regards the second question, are you aware that x^3+1 factorizes? Find a value of x that makes x^3+1 equal zero. What does that tell you about factors of x^3+1? That's enough for now. I've give you a lot of questions. Usually you don't answer them all, just ignoring all but one. I'll see what you do this time. Also, I have to go now - I'll be back tomorrow.
- pencil 16y ago
  in 1 and 3 those extra square brackets aren't necessary. secondly i'am trying to solve this as it will help in working out laplace transforms in the future. i went through the partial fraction tutorial in wikipedia and khan acadamy and i'am able to solve the following problems. 1) x/(x+1)(x-4) 2)3x+1/x^2-6x+8 it appears that the above problems are straight forward.all i have to do is solve for 'A' and 'B' after writing them as A/x+1 + B/x-4. but the problems which i have posted are not .i've tried to jiggle it around a lot of times and going no where. with regards to the second question i really don't know how on earth to factorize x^3+1 that's in the denominator. lastly i'll be really happy if you wanna test my ability in math.i'll answer your questions if i'am capable of so that you can suggest what needs to be done next. In fact i believe in learning math the hard way!!!
RiderOfGiraffes 16y ago
http://www.google.com/search?tbs=ww:1&q=partial+fractions&btnG=Search http://www.google.com/search?tbs=ww:1&q=partial+fraction...

2 points by pencil 16y ago

pencil 16y ago

i couldn't find any book that clearly explains logarithms and partial fractions nor khan acadamy was able to find a solution to this.

RiderOfGiraffes 16y ago
Did you try this one? http://www.khanacademy.org/video/introduction-to-logarithms http://www.khanacademy.org/video/introduction-to-logarithms Or this one: http://www.khanacademy.org/video/partial-fraction-expansion-1 http://www.khanacademy.org/video/partial-fraction-expansion-...
- pencil 16y ago
  ya i've tried them

ghoul2 16y ago

the second one is easy: it is given that X^2+Y^2=8XY, thus, X^2+Y^2+2XY=10XY (i.e, added 2XY to both sides) thus, (X+Y)^2 = 10XY. Taking log of both sides: 2 log (X+Y) = log 10 + log X + log Y = log 5 + log 2 + log X + log Y

ghoul2 16y ago

heh, come to think of it, so is the first one: it is given that log((a-b)/4) = log sqrta + log sqrtb, thus (a-b)/4 = sqrt(ab), thus, ((a-b)/4)^2 = ab, thus, (a-b)^2 = 16ab, thus a^2 - 2ab + b^2 = 16ab. Adding 4ab to both sides, a^2 + 2ab + b^2 = 20 ab, giving (a+b)^2 = 20ab.

pencil 16y ago
Thank you so much.what about the rest??any idea??
- ghoul2 16y ago
  the others are even easier than these. I think if you are really trying to get back into math you should not lose patience so easily and spend a bit more time trying, and don't ask for help even if it takes many days to get these questions - otherwise it will never click for you again (my own personal experience). I also hope I have not given you full solutions to homework problems you were supposed to solve by yourself. The wikipedia page on partial fractions gives you good directions on how to solve it.

RiderOfGiraffes 16y ago

What have you tried? What do you know? Rearrange 2.log(a)=3.log(b)+5 to find b in terms of a. Answering that will give us some idea of your current status the better to answer your questions.

pencil 16y ago
thank you so much for your response,actually rearragning the terms to find 'b' gives me 3logb = 2loga - 5 {upon subtracting 5 from both the sides}. this is where i need help.please look into my partial fractions problems as well if you got the time.
- RiderOfGiraffes 16y ago
  But you still don't have b as the subject of the equation. You need to get to "b=..." If you don't know how to do that, then at least tell us what you've tried. Can you make x the subject of this equation: log(x)+3=2t ? In other words, if the equation log(x)+3=2t is always satisfied, what does x equal? And I will look at the partial fractions, but I'd like to concentrate on this first. Actually, what is your definition of a logarithm? The last time we had a conversation I asked you some questions to assess your level of knowledge and you never answered them: http://news.ycombinator.com/item?id=1107996 http://news.ycombinator.com/item?id=1107996 It seems to me looking over other times you've asked questions that you claim you want to get into math, and you probably think you're doing lots of stuff, but in reality I've seen no evidence of you actually trying things. Gaining skills is a matter of spending the time. In math in particular, if you haven't spent ages getting loads of wrong answers then you won't have any real intuition as to what to do. You need loads of wrong answers, from which you can start to identify "good wrong answers". From those good wrong answers you can build an intuition as to what might work, and you can start to find excellent wrong answers. Then the moment comes when you see how to convert an excellent wrong answer into a right answer. With practice this becomes second nature, and after a time you wonder why you ever found it difficult. But until you've put in the hours of practice, it doesn't happen. People think doing math is a matter of somehow finding the right process and then following it. "Tell me what to do" is the perpetual cry. You need to try stuff. Try re-arranging the equations repeatedly into different forms until, maybe, one looks familiar. Don't just stare at stuff hoping inspiration will strike, and don't think there's a magic formula for finding the right next step. That's not how it works. So tell me, how many different ways can you rearrange the equation: log(x)+3=2t ?

RiderOfGiraffes 16y ago

1) 3x-1/[(x+2)(1-x+x^2)] 2) 1/(x^3+1) 3) 2x^2-14x+8/[(x^2+3x-2)(x-3)] In 1 and 3 are there supposed to be extra brackets? It's really hard to tell exactly what these are given that you've used no formatting. Secondly, why are you trying to solve these? Where did they come from? Solving the first one, assuming the numerator is supposed to be (3x-1), then we proceed as follows. When we add fractions like a/b + c/d the result is (ad+bc)/bd. In these questions we assume we have the result of such an addition and try to compute all the terms. Here we have a denominator of (x+2)(1-x+x^2) so we assume b=x+2 and d=1-x+x^2. that means we have the numerator formula is: (3x-1) = a(1-x+x^2) + c(x+2) In this case a and c are potentially polynomials (in fact by looking at the degrees c must be a polynomial of degree 1 larger than a). Now just try setting a and c to small polynomials, c being 1 degree larger, and compute what you get on the right hand side. Does it look like the left hand side? Probably not. What do you need to change? How can you make it better. Jiggle it around a bit and see what happens. Tell me, what if a=(x+1) and c=(x^2-2) : what does the RHS evaluate to? Why is it wrong? How can you change a and c to make it better. With regards the second question, are you aware that x^3+1 factorizes? Find a value of x that makes x^3+1 equal zero. What does that tell you about factors of x^3+1? That's enough for now. I've give you a lot of questions. Usually you don't answer them all, just ignoring all but one. I'll see what you do this time. Also, I have to go now - I'll be back tomorrow.

pencil 16y ago
in 1 and 3 those extra square brackets aren't necessary. secondly i'am trying to solve this as it will help in working out laplace transforms in the future. i went through the partial fraction tutorial in wikipedia and khan acadamy and i'am able to solve the following problems. 1) x/(x+1)(x-4) 2)3x+1/x^2-6x+8 it appears that the above problems are straight forward.all i have to do is solve for 'A' and 'B' after writing them as A/x+1 + B/x-4. but the problems which i have posted are not .i've tried to jiggle it around a lot of times and going no where. with regards to the second question i really don't know how on earth to factorize x^3+1 that's in the denominator. lastly i'll be really happy if you wanna test my ability in math.i'll answer your questions if i'am capable of so that you can suggest what needs to be done next. In fact i believe in learning math the hard way!!!

RiderOfGiraffes 16y ago

http://www.google.com/search?tbs=ww:1&q=partial+fractions&btnG=Search http://www.google.com/search?tbs=ww:1&q=partial+fraction...