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Math Help - Natural log help

  1. #1
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    Natural log help

    I am having trouble on these two problems:


    If a=2, b=3 , and c=5, evaluate the following:

    1. ln(((a^-3)(b^1))/(bc)^-2

    2. (ln(c^1))(ln(a)/(b^4))^-1


    For #1, I keep getting 15.
    So I have not been able to go on to #2 because I do not understand #1.

    Please Help!
    Thanks, lsnyder
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by lsnyder View Post
    I am having trouble on these two problems:


    If a=2, b=3 , and c=5, evaluate the following:

    1. ln(((a^-3)(b^1))/(bc)^-2

    2. (ln(c^1))(ln(a)/(b^4))^-1


    For #1, I keep getting 15.
    So I have not been able to go on to #2 because I do not understand #1.

    Please Help!
    Thanks, lsnyder
    1. Move that (bc)^{-2} into the numerator and make the exponent +2 then use the laws of logs to separate the log term (remember that b^2c^2 must be applied to both terms)

    b^2c^2 [-3ln(a)+ln(b)] = 3^25^2(-3ln(2)+ln(3)) = 225(-3ln(2)+ln(3))

    Not much point simplifying that unless you require a decimal approximation
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  3. #3
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    I keep getting that as the wrong answer.

    Its suppose to be like simple addition and subtraction.

    ln (a^-3) (b^1) - ln (bc)^-2
    ln (a^-3) (b^1) - ln (b^-2) (c^-2)
    3 ln a - ln b + 2 ln b + 2 ln c
    3 ln 2 - ln 3 + 2 ln 3 + 2 ln 5
    6 -3 +5 +7
    15

    Something like that.....
    Supposedly
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by lsnyder View Post
    I keep getting that as the wrong answer.

    Its suppose to be like simple addition and subtraction.

    ln (a^-3) (b^1) - ln (bc)^-2
    ln (a^-3) (b^1) - ln (b^-2) (c^-2)
    3 ln a - ln b + 2 ln b + 2 ln c
    3 ln 2 - ln 3 + 2 ln 3 + 2 ln 5
    6 -3 +5 +7
    15

    Something like that.....
    Supposedly
    Logarithms do not work that way, only powers of e will have rational natural logs

    Did you mean ln \left(\frac{a^{-3}b^1}{(bc)^{-2}}\right)

    If so simplify inside first: \frac{a^{-3}b^1}{(bc)^{-2}} = \frac{b^3c^2}{a^3}

    ln \left(\frac{b^3c^2}{a^3}\right) = ln(b^3) +ln(c^2) - ln(a^3) = 3ln(b)+2ln(c)-3ln(a)

    Sub in for a,b and c. That cannot be simplified more and any decimal answer will be an approximation
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