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C2 logs exam question problem

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    kinda stuck on this question , any help?



    Find the value for Y such that Log2Y= -3
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    Remember that

    logab = c

    is equivalent to

    b = ac
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    thank you !
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    (Original post by [email protected])
    kinda stuck on this question , any help?



    Find the value for Y such that Log2Y= -3
    log2y=-3
    log2y=log-3
    ylog2=log-3
    y=log-3/log2

    hope this helps
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    (Original post by dongonaeatu)
    log2y=-3
    log2y=log-3
    ylog2=log-3
    y=log-3/log2

    hope this helps
    You could confuse the OP by posting something like that, I can't tell if you're joking.

    Is it log(2y) or log2(y) as in log to the base 2? Just to note that if there's no base given, that implies base 10, which will mean your answer will be (10^-3 )/2.
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    (Original post by dongonaeatu)
    log2y=-3
    log2y=log-3
    ylog2=log-3
    y=log-3/log2

    hope this helps
    Just ignore this, OP.
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    (Original post by dongonaeatu)
    log2y=-3
    log2y=log-3
    ylog2=log-3
    y=log-3/log2

    hope this helps
    Don't mind, but your concepts of logs are very weak. I remember that you have asked a lot of question before about logs, first you need to get good at this yourself then answer others. Otherwise, you will just confuse the OP.
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    (Original post by raheem94)
    Don't mind, but your concepts of logs are very weak. I remember that you have asked a lot of question before about logs, first you need to get good at this yourself then answer others. Otherwise, you will just confuse the OP.
    what did i do wrong in that
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    (Original post by dongonaeatu)
    what did i do wrong in that
    The question is:  log_2y=-3

    We will solve it by writing it as,  2^{-3} = y

    I think i have explained this stuff to you before, right?

    You can't do this:
    log2y=-3
    log2y=log-3
    ylog2=log-3
    y=log-3/log2


    Remember if you have something like,  log_ab^c, then you can write it as,  clog_ab , but you are doing wrong.
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    (Original post by dongonaeatu)
    log2y=-3
    log2y=log-3

    (Original post by dongonaeatu)
    what did i do wrong in that

    What made you think that you could simply log the RHS
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    (Original post by TenOfThem)
    What made you think that you could simply log the RHS

    (Original post by raheem94)
    The question is:  log_2y=-3

    We will solve it by writing it as,  2^{-3} = y

    I think i have explained this stuff to you before, right?

    You can't do this:
    log2y=-3
    log2y=log-3
    ylog2=log-3
    y=log-3/log2


    Remember if you have something like,  log_ab^c, then you can write it as,  clog_ab , but you are doing wrong.
    i may have to look over logs again.
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    (Original post by dongonaeatu)
    i may have to look over logs again.
    You NEED to have a look, your concepts are still unclear.
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    (Original post by raheem94)
    You NEED to have a look, your concepts are still unclear.
    i think i'm going to fail c2
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    (Original post by dongonaeatu)
    i think i'm going to fail c2
    Don't be pessimistic, you still have time to improve.
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    dongonaeatu, you just gotta write out the 4 or 5 log rules out next to each other and rote memorise them one by one. You gotta have them all there on the same page so you don't get them mixed up.
 
 
 
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