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  1. #1
    Domokun_index's Avatar
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    Maths Question - Help me D:

    OKay so here's the dealio. I have an exam on Thursday and I've been doing past papers etc. for revision and then I came upon this question:



    I'm focusing on 9ii

    So firstly, you re-arrange loga^5 + loga you get 5loga + 0.5loga
    and logx + logy = logxy if I'm correct, so surely it should be 5.5loga^2? but apparently it's just 5.5loga
    then you divide by loga and the answer is supposed to be just 5.5
    I feel insanely stupid right now, so you can flame.

    If anyone can explain/help me out I'd be very grateful ^.^

    It's some simple shit yeah, but it's confusing the hell out of me and I have an exam Thursday so I want everything to be crystal clear.
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    Thank you ;D
    Last edited by Domokun_index; 06-04-2014 at 07:03 PM.

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    Domokun_index's Avatar
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    Hold up, is it because they're both in loga that I can add them?
    so say if it's 5logx+2logy it doesn't = 7logxy, but instead log(x^5)/(y^2)?
    but if it's 5logx+2logx it does = 7logx?

  3. #3
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    Homework section is still a thing right? Why are you posting it here?

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    Quote Originally Posted by Domokun_index View Post
    Hold up, is it because they're both in loga that I can add them?
    One mistake you made is when adding 5loga + 0.5loga and getting 5.5loga^2, as a is now an unchangeable part of the variable, rather than a simple multiplicand. Think of logA as a group in itself, you have 5 logA's + .5 logA's, and you end with 5.5 logA's. logx + logy is NOT logxy as logx and logy are unlike terms, and can not be added, while both in this ARE like terms, in logA. Remember, you are adding, not multiplying.

    Then you simply cancel the 5.5logA with the logA in the denominator, and end with a raw 5.5

    5.5logA
    ----------
    logA

    =
    5.5
    Last edited by Aborted; 06-04-2014 at 08:36 PM.
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  6. #5
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    5.5 looks correct to me, I don't know if that's your answer or not, but the math adds up.



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    Quote Originally Posted by Metamyota View Post
    Homework section is still a thing right? Why are you posting it here?
    Homework section is dead. No body ever gets a reply within a reasonable amount of time.

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    Quote Originally Posted by Doc View Post


    Homework section is dead. No body ever gets a reply within a reasonable amount of time.
    Oh, alright. Fuck man, I remember when it was actually active.

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    Quote Originally Posted by Domokun_index View Post
    OKay so here's the dealio. I have an exam on Thursday and I've been doing past papers etc. for revision and then I came upon this question: ...
    Thank me yo

    Did it on foolscap and here's the working ...

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    Domokun_index (06-05-2014),Sazyrus (06-05-2014)

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    Quote Originally Posted by Aborted View Post

    One mistake you made is when adding 5loga + 0.5loga and getting 5.5loga^2, as a is now an unchangeable part of the variable, rather than a simple multiplicand. Think of logA as a group in itself, you have 5 logA's + .5 logA's, and you end with 5.5 logA's. logx + logy is NOT logxy as logx and logy are unlike terms, and can not be added, while both in this ARE like terms, in logA. Remember, you are adding, not multiplying.

    Then you simply cancel the 5.5logA with the logA in the denominator, and end with a raw 5.5

    5.5logA
    ----------
    logA

    =
    5.5
    Thank you sir for the insight and explanation.
    My teacher kind of sucks and doesn't cover things well (he got 60.1% which is .1% into two firsts on his degree so we basically have the worst kinda teacher you can get)

    Quote Originally Posted by Avril Lavigne+ View Post


    Thank me yo

    Did it on foolscap and here's the working ...
    That's a good way of looking at it, thank you ^.^

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