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    Can anyone help me do part a for question 8 and 9? I don't even know where to start the question. I have just started doing logs and I don't know much about it! Any good tips would be useful as well!!
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    log a x = log a 5 + 2 log a 3

    use laws of logs to bring the 3 up to the power

    log a x = log a 5 + log a 3^2

    Then use laws of logs to put the logs together

    log a x = log a (5*3^2)
    log a x = log a 5*9
    log a x = log a 45

    so x=45
    for part 8 a
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    For part A of Q9, you can do the same once you've expanded the brackets, you can use the same laws, but remember that log a x - log a y = log a (x/y)
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    In question 9 you need to use 2 laws of logs - the one where "log(a) - log (b) = log(a/b), and the one where "n log (a) = log (a^n).

    So you can simplify the bit in the brackets to 2 log (k/2), then you can move the 2 to being the power using the second rule so it is now log (k/2)^2. This can be expended out to log ((k^2)/4).

    Comparing the equations now you know that x = (k^2)/4, so that can be rearranged to give 4x = k^2.


    I hope that's clear and the notation isn't too confusing. Being familiar with the log laws is the most useful thing you can do so make sure you know them
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    (Original post by LRJ)
    log a x = log a 5 + 2 log a 3

    use laws of logs to bring the 3 up to the power

    log a x = log a 5 + log a 3^2

    Then use laws of logs to put the logs together

    log a x = log a (5*3^2)
    log a x = log a 5*9
    log a x = log a 45

    so x=45
    for part 8 a
    Thanks that's the correct answer!! but how do you 'bring the 3 up to the power'? Can you please tell me which law you used?

    Also at the end (log a x = log a 45) did you just cancel the logs?

    Thanks once again
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    (Original post by fathimadeen)
    Thanks that's the correct answer!! but how do you 'bring the 3 up to the power'? Can you please tell me which law you used?

    Also at the end (log a x = log a 45) did you just cancel the logs?

    Thanks once again
    No prob - it's a bit late to do maths isn't it? xD

    The law of log you need to bring the 3 up to the power is:
    k log a x = log a x^k

    And at the end, (log a x = log a 45), it's not cancelling logs, but just that A can only be 45 since the rest of the equation is exactly the same on both sides

    It's kinda like equating the coefficients, i.e if Ax + B = 3x + 5
    A must be 3, and B must be 5
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    (Original post by LRJ)
    No prob - it's a bit late to do maths isn't it? xD

    The law of log you need to bring the 3 up to the power is:
    k log a x = log a x^k

    And at the end, (log a x = log a 45), it's not cancelling logs, but just that A can only be 45 since the rest of the equation is exactly the same on both sides

    It's kinda like equating the coefficients, i.e if Ax + B = 3x + 5
    A must be 3, and B must be 5
    Right I get it now!! hehe yea it is kinda late but got tonnes of homework to do so have to stay awake and finish them all xx
 
 
 
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