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Trouble with logarithms and partial fractions

2 points by pencil ↗ HN
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)]

49 comments

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ghoul2
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
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
Thank you so much.what about the rest??any idea??
ghoul2
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
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
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
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

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 ?

pencil
i completely agree with you sir and i really appologise for not responding to this thread http://news.ycombinator.com/item?id=1107996 to be honest with you i really don't know where i stand .what's my level of competence. i recently gained interest in math and physics because i regret not taking these two subjects seriously back in college.now i work for a bank which pay's peanuts.i don't intend to get a job which helps me build my math/physics skills,i just wanna learn it because i have a burning desire to learn it.it's as simple as that. yes.math requires hours and days of practice which i'am religiously doing but it's not enough for me as it's very evident by the level of knowledge/intellect that i possess. with that being said the number of ways arranging the equation log(x)+3=2t hmm.. log(x)=2t-3 am i right??hope i'am not bullshitting you.

by the way the only definition of logarithm that i know is 'log is inverse of exponents'

RiderOfGiraffes
So what does that mean?

OK, here are some terms to simplify. In each case, what is a simpler way to write the expression:

  x^3 * x^5

  (t^3)^2

  u^(3^2)
Expand the following:

  (g+2)^2

  (z-3)^3
Now, if x=10^y, what does y equal?
pencil
x^3+5=x^8

t^3 * t^3 = t^6

i guess it's u^6

g^2+4g+4

not sure abt (z-3)^3

expanding (z-3)^3

could it be: (z-3).(z+3)^2 ??

=>(z-3) (z^2+3z+9)

RiderOfGiraffes
OK, in the first one you need brackets : x^3+5 is not usually understood to be the same as x^(3+5), but the answer is right, and the process is right. The others are right too.

So (g+2)^2 is equal to g^2+4g+4, but it's also equal to (g+2).(g+2) [where I've used the dot for multiplication]

Does that help you for (z-3)^3 ??

pencil
now i have the solution for y in x=10^y

here it is:x=10^y

subtract 10^y from both the sides

x-10^y=0

now subtract x from both the sides

-10^y = -x

divide by -1

-10^y/-1 = -x/-1

= 10^y = x

take log base 10 both sides

log10^y = logx

= ylog10=logx

y = logx/log10

RiderOfGiraffes
Ah, very good. You've gone slightly the long way around, but you've got the right answer.

The first part you go from x=10^y to -10^y=-x to 10^y=x. Is it not clear that if x=10^y then 10^y=x? The point is that any time A=B, then B=A.

Next, you have ended with y=log(x)/log(10). If the logarithms are base 10, what is log(10)? Can you use that to make your result simpler?

Next question: if t=log(u) (where the logarithm is base 10) then what does u equal?

pencil
i'll be back in 45 min
pencil
i'am not able to solve t=log(u).not even a single step sorry

i'am assuming this would end up in some number called 'e' which i came accross somewhere in the past.please help..i'am not sure what that number means and how to get it

RiderOfGiraffes
OK, I really don't know how to help you now using this medium. You claim to have watching the Khan Academy video, but this question is answered in that.

The simple rules are these:

    log(a.b) = log(a) + log(b)

  hence

    log(a^4) = log(a.a.a.a) = log(a)+log(a)+log(a)+log(a) = 4.log(a)

  Therefore

    if  a = b^z
    then log(a) = log(b^z)
    so   log(a) = z.log(b)
This is true for all bases. Then we have

    log_b(b) = 1  :  Log (base b) of b is 1.

    Hence log_b(b^2) = 2
          log_b(b^3) = 3

    and so on.
In all this the basic rule is this:

    if   x = b^y
    then log_b(x) = log_b(b^y)
                  = y.log_b(b)
                  = y
So if the base is 10, then we have:

    If  a = 10^b
    then log(a) = log(10^b)
                = b.log(10)
                = b.1
                = b
So now, start with the equation t=log(u), assuming the log is base 10, and use the above rules to change it around. When you do that, see if you can get to saying u equals something.

Finally, you need to study sections 1, 2 and 3 of this page:

http://en.wikipedia.org/wiki/Logarithm

What's on that page will also help you to understand the questions.

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RiderOfGiraffes
Let's go much simpler, right back to logs basics.

If y=log(x) (with 10 as the base of the logarithms) then what does x equal in terms of y?

pencil
i'am assuming this would end up in some number called 'e' which i came accross somewhere in the past.please help..i'am not sure what that number means and how to get it
RiderOfGiraffes
No, it doesn't. I've given you the rules, and I've pointed you at two web pages that explain the rules and work through some examples. You seem to keep wanting me to just give you the answers without showing me any working for yourself.

This is not the medium for teaching you about the basic, basic, basic rules of elementary algebra. There are loads of tutorials on the web - try using Google to search for tutorials on logarithms.

When you tell me what you've searched for, and what the first rules are that they give you, then I'll start helping again.

RiderOfGiraffes
I found you a tutorial:

http://www.phon.ucl.ac.uk/cgi-bin/wtutor?tutorial=t-log

Work through that - it will help with your basic understanding of what's going on.

pencil
one of the best tutorials on logs i've ever come accross.thanks a million!!!! please look into my partial fraction problems when you get the time.so far i've been able to solve a couple of them but the above mentioned one is bothering me a lot.
RiderOfGiraffes
It was the very, very first one I found with a Google search for "logarithm tutorial".

Have you not tried searching for tutorials?

RiderOfGiraffes 

   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
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
You've got real problems with brackets. Major problems that are going to cause significant difficulties down the line. I'll bet the square brackets are necessary, and you need more brackets for the numerator.

Your number 1 above you quote as:

  x/(x+1)(x-4)
I'll bet you don't mean that. When you have multiplication and division at the same level, you evaluate left-to-right. The above is the same as:

  [ x / (x+1) ] (x-4)
which is the same as:

  x (x-4) / (x+1)
which I'll bet is not what you intended.

The second - as you quote it - is:

    3x+1/x^2-6x+8
Division binds more closely than addition and subtraction, while addition and substraction are then read from left-to-right. What you quote is therefore the same as:

    3x + 1/x^2 - 6x + 8
which is

    3x + (1/x^2) - 6x + 8
which simplifies to

    (1/x^2) - 3x + 8
I'll bet that's not what you meant.

Now let's turn to the first of the ones you originally asked about:

    3x-1/[(x+2)(1-x+x^2)]
I'll bet you mean:

    (3x-1) / [(x+2)(1-x+x^2)]
and I'll bet the square brackets are necessary. And you are right that all you need to do is solve for A and B after writing it as A/(x+2) + B/(1-x+x^2).

So remembering that a/b + c/d = (ad+bc)/bd, what does

    A/(x+2) + B/(1-x+x^2)
equal?
pencil
all right the brackets concept makes sense.

remembering a/b + c/d = (ad+bc)/bd

A/(x+2) + B/(1-x+x^2) = A(1-x+x^2) + B(x+2)/(x+2)(1-x+x^2)

correct??

RiderOfGiraffes
Yes, except you forgot the brackets again. You should have:

    [ A(1-x+x^2) + B(x+2) ] / [ (x+2)(1-x+x^2) ]
You probably knew that, but this is really, really important. Not getting it right now will confuse you a great deal later.

However, now you have this as your numerator:

    A(1-x+x^2) + B(x+2)  [Eqn *]
Look again at your original problem

- what do you want the numerator to be?

- Expand out Eqn * to remove the brackets.

- What values can you assign to A and B to make Eqn * be the numerator you want?

As a hint, A and B are not simply numbers.

pencil
first of all i don't know the significance of replacing the numerator with capital letters when decomposing partial fractions.i'am not aware of the practical implications of these problems.all i know is it'll come in handy when solving laplace transforms in the future which is used in physics/electrical etc.(that's what i'am after). and if you are curious to know why i wanna learn these..well i don't have a proper reason.i simply want to!!

i looked back at the original problem but i'am unable to figure out the values to assign to A and B.

RiderOfGiraffes
OK, to recap. You want to decompose this:

    [ 3x-1 ] / [ (x+2)(1-x+x^2) ]
into fractions. In other words, you want to find A and B such that:

    A/(x+2) + B/(1-x+x^2) = [ 3x-1 ] / [ (x+2)(1-x+x^2) ]
Apart from getting the brackets wrong, you said (correctly) that the left hand side is equal to this:

    [ A(1-x+x^2) + B(x+2) ] / [ (x+2)(1-x+x^2) ]
So now you need to solve:

[ A(1-x+x^2)+B(x+2) ] / [ (x+2)(1-x+x^2) ] = (3x-1) / [ (x+2)(1-x+x^2) ]

The denominators are equal, so you just need to make the numerators equal.

So tell me - what equation do you have to solve?

pencil
i get A(1-x+x^2) + B(x+2) = 3x-1
RiderOfGiraffes
Excellent.

+ Expand B(x+2)

+ Expand A(1-x+x^2)

What do you get? You need to play with these sorts of equations and see what happens - see what you get.

pencil
expand B(x+2)?? but how??
RiderOfGiraffes
How would you expand 3(x+2)?
pencil
3x+6
RiderOfGiraffes
Or, before simplifying, 3x+3.2 (using a dot for multiplication)

So how do you expand B(x+2) ?

pencil
Bx+B2 ??? now it's getting funny!!!!!!!!
RiderOfGiraffes
So going back, you need to solve: [ A(1-x+x^2)+B(x+2) ] / [ (x+2)(1-x+x^2) ] = (3x-1) / [ (x+2)(1-x+x^2) ]

As I said, the denominators are already equal, so you need to make the numerators equal.

That means you need to make this:

    A(1-x+x^2)+B(x+2)
equal to (3x-1). Expand this - you already know that the second part becomes B.x+2B - and have a look.

Then think - what can you set A and B to be to make it equal to 3x-1?

What if you set A to be x?

What if you set B to be 2?

What if you set A to be -3?

What if you set B to be 2x?

Try those, try more, see if you can work out how you can get 3x-1. Here's a hint: when you expand you'll get one of the terms as Ax^2. How can you get that to cancel out?

Try things. Tell me what you see, tell me what you get.

pencil
i've started out with calculus.(it;s been 30 hours now) and i'am enjoying it as i'am able to understand.so my plan is to continue this partial fraction problem when i reach integral calculus involving partial fractions and eventually laplace transforms. and you said my algebra skills sucks.well i partialy agree with you cause i wouldn't be able to solve limits and differentiate functions without a grasp on algebra(precalculus for namesake). as a matter of fact i've never been good at anything .i suck in everything that i do.!!!!!!!!!
RiderOfGiraffes
Just to add another question - why is that funny? A and B represent polynomials - why should this be funny to you?
pencil
the fun part has got nothing to do with polynomials.i'am finding it funny cause i'am taking ages to learn this basic concept .to put in better sentence a DUMB ASS(that's me) taking advice from a knowledgeable person(that's you)

by the way thanks very much for telling me that 'A' and 'B' represents polynomials.never knew it before.

RiderOfGiraffes
It's not necessarily that you're dumb - I suspect you aren't. The problem is that you're trying to get a grip on these techniques when you haven't really mastered the absolute basics of algebra.

It's not surprising - you're trying to teach yourself, and it's always going to be that there are gaps. You need properly constructed tutorials and exercises.

Have a look at these:

http://www.google.co.uk/#q=algebra+tutorial+with+exercises

Pick one of the tutorials and work through it completely. Then you can do a google search for partial fraction tutorials and do the same.

pencil
i've started out with calculus.(it;s been 30 hours now) and i'am enjoying it as i'am able to understand.so my plan is to continue this partial fraction problem when i reach integral calculus involving partial fractions and eventually laplace transforms. and you said my algebra skills sucks.well i partialy agree with you cause i wouldn't be able to solve limits and differentiate functions without a grasp on algebra(precalculus for namesake). as a matter of fact i've never been good at anything .i suck in everything that i do.!!!!!!!!!

by the way i'am not offended.i'am enjoying myself with calculus.i'll get back to you if i have any doubts.if you got to say something you can.

[deleted]
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RiderOfGiraffes
I didn't say your algebra skills suck - I say there are gaps. You don't put brackets where you should, and you don't seem to be able to manipulate equations quickly and easily. But you have some skills, and if you're only just starting out then you're probably doing well. I can't tell from this distance via this medium.

As a final hint, here's something interesting:

  For a polynomial p(x), if p(a)=0 for some value a,
  then (x-a) is a factor of p(x).

  That means, for example, that (x+1) is a factor of
  x^3+1 and (x-1) is a factor of x^3-1.
I have to go. Good luck!