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Math Help - Logarithms

  1. #1
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    Post Logarithms

    Hi,
    I'm not sure if I got these problems right. Can someone help me?

    1. write "ln5" as a multiple of a common logarithm.
    2. write "ln(x^2-1)/x^3)" as a sum, difference and/or constant multiples of logarithms. I got "ln(x^2-1)-3lnx" for this.
    3. Simplify "log(9/300)". I got "log3"
    4. Simplify -8+e^(ln(x^3)). I have no idea how to solve this one

    I'll really appreciate it if someone can answer these questions. Thanks!

    Isabel
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  2. #2
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    Quote Originally Posted by cuteisa89 View Post
    Hi,
    I'm not sure if I got these problems right. Can someone help me?

    1. write "ln5" as a multiple of a common logarithm.
    ln 5 = log 5/ln e
    2. write "ln(x^2-1)/x^3)" as a sum, difference and/or constant multiples of logarithms. I got "ln(x^2-1)-3lnx" for this.
    ln(x^2-1) - ln (x^3) = ln [(x+1)(x-1)]- 3ln x=ln(x+1)+ln(x-1)-3*ln x
    3. Simplify "log(9/300)". I got "log3"
    9/300=300
    Thus,
    log 9/300 = log 300 = log 3*10^2 = log 3 + log 10^2=log 3+2
    4. Simplify -8+e^(ln(x^3)). I have no idea how to solve this one
    Note
    e^(ln x)=x for all positive x.
    Now,
    ln x^3 = 3ln x
    Thus,
    e^(ln(x^3))=e^(3ln x)=(e^(ln x))^3 (by the exponent laws).
    But then,
    (e^ln(x))^3=x^3.

    Or, an easier way.
    Let y=x^3.
    Thus,
    e^(ln y) = y= x^3.

    This is Mine 5-th Post!!!
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cuteisa89 View Post
    1. write "ln5" as a multiple of a common logarithm.
    i'm not exactly sure what they expect here.

    2. write "ln(x^2-1)/x^3)" as a sum, difference and/or constant multiples of logarithms. I got "ln(x^2-1)-3lnx" for this.
    you are right so far, but you can go further:

    ln(x^2-1)-3lnx = ln(x + 1)(x - 1) - 3lnx = ln(x + 1) + ln(x - 1) - 3lnx

    3. Simplify "log(9/300)". I got "log3"
    is this log to the base 3 or the normal log to the base 10? (either way, log3 is wrong)

    log(9/300) = log(3/100) = log3 - log100 = log3 - 2 (if it's log to the base 10)


    4. Simplify -8+e^(ln(x^3)). I have no idea how to solve this one
    -8+e^(ln(x^3)) = -8 + x^3 = x^3 - 8


    note that e^ln(x) = x
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    ln 5 = log 5/ln e

    ln(x^2-1) - ln (x^3) = ln [(x+1)(x-1)]- 3ln x=ln(x+1)+ln(x-1)-3*ln x

    9/300=300
    Thus,
    log 9/300 = log 300 = log 3*10^2 = log 3 + log 10^2=log 3+2

    Note
    e^(ln x)=x for all positive x.
    Now,
    ln x^3 = 3ln x
    Thus,
    e^(ln(x^3))=e^(3ln x)=(e^(ln x))^3 (by the exponent laws).
    But then,
    (e^ln(x))^3=x^3.

    Or, an easier way.
    Let y=x^3.
    Thus,
    e^(ln y) = y= x^3.

    This is Mine 5-th Post!!!
    now i know how Soroban feels when he replies to a question only to see I beat him to it
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