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Math Help - Why is it I can never find those rules explained

  1. #1
    Newbie bbiandov's Avatar
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    Why is it I can never find those rules explained

    Why is it I can never find those rules explained. I was able to find ton of sites explaining the different simple rules of moving things around so that the variable is at one side and the numbers on the opposite side of the equal sign but NEVER did I see this simple rule explained?

    Yes this is not the whole story as the last step is to ln both sides and so forth but I really wanted to see the a/b=c explained.

    Am I missing something or this is sooo retareded so no one bothers to explain it?

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  2. #2
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    You did it wrong. 0.25 is not equal to its reciprocal. And you don't need to arrange it so that the variable on one side and the value is on the other, if it's too much hassle. Why? If you know these two properties:

    \ln\left(\frac{A}{B}\right) = \ln{A} - \ln{B}

    \ln{1} = 0
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  3. #3
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    Quote Originally Posted by bbiandov View Post
    Why is it I can never find those rules explained. I was able to find ton of sites explaining the different simple rules of moving things around so that the variable is at one side and the numbers on the opposite side of the equal sign but NEVER did I see this simple rule explained?

    Yes this is not the whole story as the last step is to ln both sides and so forth but I really wanted to see the a/b=c explained.

    Am I missing something or this is sooo retareded so no one bothers to explain it?

    Believe it or not, I do not fully understand what you mean.
    The "ln" thing confuses me more.

    If I can follow your question as related to the examples shown, then my understanding would be that you want to know why,

    in a/b = c,

    Case 1..... a/c = b ?
    and
    Case 2 .... a = b*c ?

    Okay, in Case 1,
    a/b = c
    Clear the fraction, multiply both sides by b,
    a = b*c
    So, solving for b, divide both sides by c
    a/c = b

    In Case 2,
    a/b = c
    Clear the fraction, multiply both sides by b,
    a = b*c
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  4. #4
    Newbie bbiandov's Avatar
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    Thank you. I see the sequence of clearing the fractions now.

    The ln business was mentioned because I really need to find "t" which is the exponent but until I get to a point where each side is simplified I treat the whole number with "t" exponent as simply X

    Once I am down to simple notation I then replace X with 1.015^t and ln both sides so that ^t can be moved in front and then solve for t

    So to summarize - to clear a fraction you multiply both sides by the Denominator of the fraction that needs to be cleared? Whether Denominator is variable or value is not relevant to this decision?

    ~B
    Last edited by bbiandov; September 1st 2008 at 09:23 PM.
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  5. #5
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    Quote Originally Posted by bbiandov View Post
    Thank you. I see the sequence of clearing the fractions now.

    So to summarize - to clear a fraction you multiply both sides by the Denominator of the fraction that needs to be cleared? Whether Denominator is variable or value is not relevant to this decision?

    ~B
    Yes.

    In Algebra, we use variables or constants or values as if they are numbers.

    (There are cases where you need to clear all fractions....if there are more than one fraction in the equation.....)
    But before this might confuse you, let's leave it at what you have "learned" above so far.
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  6. #6
    Newbie bbiandov's Avatar
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    Thanks ticbol

    Now I am sure I could have found those rules posted in an FAQ? While I really appreciate your answering such low level questions here I am curious - where is the FAQ?

    I want to see the rest of the rules - specifically the rules for moving things back and forth across the equal sign and reversing their sign. Say you move 5x across and it has to become -5x?

    Thanks
    ~B
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  7. #7
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    Quote Originally Posted by bbiandov View Post
    Thanks ticbol

    Now I am sure I could have found those rules posted in an FAQ? While I really appreciate your answering such low level questions here I am curious - where is the FAQ?

    I want to see the rest of the rules - specifically the rules for moving things back and forth across the equal sign and reversing their sign. Say you move 5x across and it has to become -5x?

    Thanks
    ~B
    Teaching is teaching, whether it is for low level or high level, as long as the "student" wants to learn.
    I am more interested in "teaching" in the low levels because that's were foundations are built. Just like in constructions, a good foundation in Math will give easier or more stable "super structure" or understanding in higher Math.

    (I am not a teacher. I am in constructions. I like more to teach laborers how to become skilled workers than to teach skilled workers new construction methods or ways....although I do both and "in -betweens". Like I am more happy to guide a tot learning how to walk than to guide a brat learning how to slide his board on a handrail or on street curbs.)

    I don't know about a FAQ too in this Forum, but if it is about transferring back and forth on both sides of the equation, then here is additional thing for you.

    The 5x becoming (-5x) when transposed to the other side?
    Well, yes, anything transposed to the other sides of the equation will just change its sign when transposed.
    The 5x will become (-5x). The (-m^2) will become m^2. That is automatic after you're used to it.

    The reason is based on Math, of course.
    3 -5x = m^2 ------(i)

    If you want to transfer or tranpose the 5x to the RHS,
    3 = m^2 +5x

    That is because
    3 -5x = m^2
    Add 5x from both sides,
    3 -5x +5x = m^2 +5x
    3 = m^2 +5x ---------see?

    If you want to transfer the m^2 in Eq.(i) to the other sde,
    3 -5x = m^2
    3 -5x -m^2 = 0 ------that is it. Easy.

    That is because
    3 -5x = m^2
    Subtract m^2 from both sides,
    3 -5x -m^2 = m^2 -m^2
    3 -5x -m^2 = 0 ------see again?


    These explanations are more effective, or easier to explain and/or easier to understand, if done in person-to-person or where teacher and learner are both present in persons....where there will be eye-contacts and easier "meetings of minds". Back and forth asking and answering is done quickly, if necessary.

    In Forum or online meetings like this, it is left to chance mostly. Chances are the teacher has explained it in a manner that the learner understant it, or chances are the learner is more confused after the explanation. That's why the back and forth asking and answering between the poster and the replier could go on for months. Or, the poster just shrug his shoulder and stop asking any further because the replier might lose his temper. . Or the poster might appear to be really very poor in Math if he would continue asking and asking on the same question even if the replier keeps on answering and answering.....the poster might as well quit asking even if he is not cleared yet by the replier's thousand replies on the same question.
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  8. #8
    Newbie bbiandov's Avatar
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    Thank you

    I just got stuck at something interesting:

    80x+240x^3=1,000,000

    It doesnt fit quadratic formula becaue of ^3

    So then how does one solve this?

    ~B
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  9. #9
    A riddle wrapped in an enigma
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    Quote Originally Posted by bbiandov View Post
    Thank you

    I just got stuck at something interesting:

    80x+240x^3=1,000,000

    It doesnt fit quadratic formula becaue of ^3

    So then how does one solve this?

    ~B
    Not easily. You might play around with the cubic formula , but that's complicated. I used my TI-84 to approximate the only real root of 16.084585.
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by bbiandov View Post
    Thank you

    I just got stuck at something interesting:

    80x+240x^3=1,000,000

    It doesnt fit quadratic formula becaue of ^3

    So then how does one solve this?

    ~B
    1. remove any common factors and rearrange to taste, I like:

    3x^3+x-12500=0

    Now we could use the cubic formula (see masters' post), but that is seldom done. However this is a depressed cubic and can be reduces to a quadratic in t^3 by the substitution t=x-1/(9x), which can then be solved for t using the quadratic formula, and then we will have a number of other quadratics to solve to get x

    Using Descartes rule of signs we see that this has 1 positive root, and zero negative roots. Which narrows things down a bit.

    Now by plotting the relation:

    y=3x^3+x-12500

    we can find that the root is close to 16, which you can refine by numerical means (trying out values near 16 to locate the root more precisely will be good enough for some purposes)

    RonL
    Last edited by CaptainBlack; October 4th 2008 at 12:59 AM.
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