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    Hey guys if anyone out there can explain the answer to a question for me that would be great!

    The question is Q9 ii. (June 2009 mei c2)

    It is a logs question but ill have a go at typing it out:

    (Loga^5+Loga^1/2)/Loga

    So what i did was bring the powers to the front and did this

    5Loga+0.5Loga-1Loga

    then just did 5+0.5-1= 4.5

    However looking at it I can see you just need to do (5+0.5)/1 :/

    so my question is: why do we not need to use log rules to make the equation Loga5+Loga0.5-Loga? (that is what our teacher taught us!)

    If i haven't explained very well please tell me
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    (Original post by Davelittle)
    Hey guys if anyone out there can explain the answer to a question for me that would be great!

    The question is Q9 ii. (June 2009 mei c2)

    It is a logs question but ill have a go at typing it out:

    (Loga^5+Loga^1/2)/Loga

    So what i did was bring the powers to the front and did this

    5Loga+0.5Loga-1Loga

    then just did 5+0.5-1= 4.5

    However looking at it I can see you just need to do (5+0.5)/1 :/

    so my question is: why do we not need to use log rules to make the equation Loga5+Loga0.5-Loga? (that is what our teacher taught us!)

    If i haven't explained very well please tell me
    Because you have something that looks like (k log a) / log a which is just k (in this case k = 5.5)

    You would use -log a if you had a term that said log(1/a) because 1/a is the same as a^-1, but you don't have that here - 1/loga is NOT the same as log(1/a).
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    did this question quickly, see if it helps?
    Name:  IMG_20130515_192818_305[1].jpg
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    sorry for my hand writing

    bassically the log a's cancel, not take away

    I guess you could imagine them going 5log(a-a) + 1/2log (a-a) which would be (5*1) + (1/2*1)
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    (Original post by davros)
    Because you have something that looks like (k log a) / log a which is just k (in this case k = 5.5)

    You would use -log a if you had a term that said log(1/a) because 1/a is the same as a^-1, but you don't have that here - 1/loga is NOT the same as log(1/a).
    ok thanks i understand!

    i swear my teacher told us that subtracting logs is the same as dividing logs and dividing is the same as subtracting like this. loga-logb=loga/b
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    I think that is true, but loga/logb /=/ log a - log b
    i think it's only log(a/b)= loga - logb?
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    (Original post by Davelittle)
    ok thanks i understand!

    i swear my teacher told us that subtracting logs is the same as dividing logs and dividing is the same as subtracting like this. loga-logb=loga/b
    That's true, but as I tried to point out (perhaps not clearly) you have one log divided by another log, NOT the log of (one thing divided by another).
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    (Original post by davros)
    That's true, but as I tried to point out (perhaps not clearly) you have one log divided by another log, NOT the log of (one thing divided by another).
    ah because its a^5 and a?

    thanks!!!!!
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    no, not because it's A^5 divided by A, it's because it's log A/Log A instead of Log (A/A)
 
 
 
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