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Math Help - Urgent help with logs needed!!!

  1. #1
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    Exclamation Urgent help with logs needed!!!

    Hey, the question is:

    7(i) Express each of the following in terms of log_10 x and log_10 y.

    (a) log_10 (x/y)

    (b) log_10 (10x^2y)

    (ii) Given that

    2 log_10 (x/y)= 1+ log_10 (10x^2y)

    find the value of y correct to 3 decimal places.

    If someone could help me with this i'd be most grateful!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by LoveDeathCab View Post
    Hey, the question is:

    7(i) Express each of the following in terms of log_10 x and log_10 y.

    (a) log_10 (x/y)

    (b) log_10 (10x^2y)

    (ii) Given that

    2 log_10 (x/y)= 1+ log_10 (10x^2y)

    find the value of y correct to 3 decimal places.

    If someone could help me with this i'd be most grateful!
    Laws of logs:

    log_c(a) = b means that c^b=a

    log(a.b) = log(a) + log(b)

    log(a/b) = log(a) - log(b)

    log(a^b) = b log(a).

    So:

    (a) log_10 (x/y) = log_10(x) - log_10(y)

    (b) log_10 (10x^2y) = log_10(10) + log_10(x^2 y) = log_10(10) + log_10(x^2) + log_10(y)

    ........................... = log_10(10) + 2 log_10(x) + log_10(y)

    Now from the definition of logs log_a(a) = 1, so log_10(10)=1, so:

    log_10 (10x^2y) = 1 + 2 log_10(x) + log_10(y)


    (ii) Given that

    2 log_10 (x/y)= 1+ log_10 (10x^2y)

    using part (a) above on the left hand side, and (b) on the right hand side:

    2 [log_10(x) - log_10(y)] = 1+1 + 2 log_10(x) + log_10(y) = 2 + 2 log_10(x) + log_10(y),

    so:

    -3 log_10(y) = 2

    or: log_10(y) = -2/3, so y=10^(-2/3) or : y ~=0.215.

    RonL
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