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    Log question - express in terms of logax , logay, logaz

    Ive attached the ques h.) & my working but i dont understand where the 5/2 comes from.

    Answer is 5/2 + 1/2logax - 3logay + 5/2logaz

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    (Original post by kiiten)
    Log question - express in terms of logax , logay, logaz

    Ive attached the ques h.) & my working but i dont understand where the 5/2 comes from.

    Answer is 5/2 + 1/2logax - 3logay + 5/2logaz

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    \log_2 4 = \log_2 2^2 = 2\log_2 2 = 2 \times 1 = 2

    \displaystyle \log_a \frac{\sqrt{axz^5}}{y^3} = \log_a \sqrt{axz^5} - \log_a y^3

    Now \log_a y^3 = 3\log_a y

    and \log_a \sqrt{axz^5} = \frac{1}{2} \log_a (axz^5) = \frac{1}{2}\left(\log_a a + \log_a x + \log_a z^5\right)

    Can you take it from here? What's \log_a a?
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    (Original post by Zacken)
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    \log_2 4 = \log_2 2^2 = 2\log_2 2 = 2 \times 1 = 2

    \displaystyle \log_a \frac{\sqrt{axz^5}}{y^3} = \log_a \sqrt{axz^5} - \log_a y^3

    Now \log_a y^3 = 3\log_a y

    and \log_a \sqrt{axz^5} = \frac{1}{2} \log_a (axz^5) = \frac{1}{2}\left(\log_a a + \log_a x + \log_a z^5\right)

    Can you take it from here? What's \log_a a?
    How did you get to -logay^3 on the second line of working ?
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    (Original post by kiiten)
    How did you get to -logay^3 ?
    \log \frac{a}{b} = \log a - \log b except in this case, we have a =\sqrt{axz^5} and b = y^3
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    (Original post by Zacken)
    \log \frac{a}{b} = \log a - \log b except in this case, we have a =\sqrt{axz^5} and b = y^3
    Umm i think ive done it wrong but i got
    2 + 1/2loga + 1/2logx + 5/2logz - 3logay
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    (Original post by kiiten)
    Umm i think ive done it wrong but i got
    2 + 1/2loga + 1/2logx + 5/2logz - 3logay
    Yeah, that's correct, just remember that \frac{1}{2} \log_a a  = \frac{1}{2} \times 1 = \frac{1}{2} so you can simplify your first two terms as 2 + \frac{1}{2} = \cdots
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    (Original post by Zacken)
    Yeah, that's correct, just remember that \frac{1}{2} \log_a a  = \frac{1}{2} \times 1 = \frac{1}{2} so you can simplify your first two terms as 2 + \frac{1}{2} = \cdots
    So 5/2 + 1/2logax + 5/2logaz - 3logay
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    (Original post by kiiten)
    So 5/2 + 1/2logax + 5/2logaz - 3logay
    Seems fine, yups.
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    Thank you

    If you have log (1+5^-x) would it simplify to -xlog6 ?
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    (Original post by kiiten)
    Thank you

    If you have log (1+5^-x) would it simplify to -xlog6 ?
    No, it wouldn't; you can't simplify that anymore. Remember that \log(a+b) \neq \log a+ \log b. :-)
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    (Original post by Zacken)
    No, it wouldn't; you can't simplify that anymore. Remember that \log(a+b) \neq \log a+ \log b. :-)
    So how would you solve
    Log (1×5^-x) = log3
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    (Original post by kiiten)
    So how would you solve
    Log (1×5^-x) = log3
    "cancel" the logs: 1 + 5^{-x} = 3 \Rightarrow 5^{-x} = 2 \Rightarrow 5^x = \frac{1}{2} - now you know how to solve this using logarithms.
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    (Original post by Zacken)
    "cancel" the logs: 1 + 5^{-x} = 3 \Rightarrow 5^{-x} = 2 \Rightarrow 5^x = \frac{1}{2} - now you know how to solve this using logarithms.
    Oh i see - could you do that for any log e.g. log5^-x = log3 instead of simplifying
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    (Original post by kiiten)
    Oh i see - could you do that for any log e.g. log5^-x = log3 instead of simplifying
    If you have \log a = \log b - then you know that a=b.

    If you have \log a + \log b = \log c IT IS NOT TRUE THAT a+b = c BUT IT IS TRUE that \log a + \log b = \log ab = \log c \Rightarrow ab=c

    Basically when you have log equations you should try and get it in the form \log (\text{something}) = \log (\text{something else}) then you can say something = something else. (the bases of the logs have to be the same, if they're not, you can always use change of base rule) .
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    (Original post by Zacken)
    If you have \log a = \log b - then you know that a=b.

    If you have \log a + \log b = \log c IT IS NOT TRUE THAT a+b = c BUT IT IS TRUE that \log a + \log b = \log ab = \log c \Rightarrow ab=c

    Basically when you have log equations you should try and get it in the form \log (\text{something}) = \log (\text{something else}) then you can say something = something else. (the bases of the logs have to be the same, if they're not, you can always use change of base rule) .
    Thanks that helps to explain things
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    (Original post by kiiten)
    Thanks that helps to explain things
    No problem.
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    What about this question - im not sure where you would start. Ive attached some workingAttachment 513103513107

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    Use Log rules.
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    (Original post by Mystery.)
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    Use Log rules.
    Have i done it right so far. So, 5/ (log3x) = 6 / (log3)
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    (Original post by kiiten)
    What about this question - im not sure where you would start. Ive attached some workingAttachment 513103513107

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Views: 118
Size:  498.4 KB
    You have

    \log_3 x + log_3 y = 5

    \log_3 x \times log_3 y = 6


    Let a=\log_3 x and b=\log_3 y then you have

    a+b=5

    ab = 6

    So you can ignore logs and solve these equations for a and b. Then substitute back in for a and b.
 
 
 

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