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Math Help - Logarithms...please help

  1. #1
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    Logarithms...please help

    I am having a very hard time with these...I am very confused and unsure of where to start with these...if someone could please guide me through one or two of them to help me understand, i would greatly appreciate it....

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    Quote Originally Posted by Raleigh View Post
    I am having a very hard time with these...I am very confused and unsure of where to start with these...if someone could please guide me through one or two of them to help me understand, i would greatly appreciate it....

    hello,

    to #9:

    use the base 8 at both sides of the equation:

    8^{\log_8{(4x+14)}}=8^2~\implies~4x+14=64 . Please continue.

    to #10:

    6 \cdot e^{0.05 \cdot x} + 29 = 83~\iff~e^{0.05 \cdot x}=9~\implies~0.05 \cdot x = \ln(9) . Please continue.

    to #11:

    \frac{400}{1+e^{-x}}=375~\iff~400=375+375 \cdot e^{-x}~\iff~\frac1{15}=\frac1{e^x}~\implies~e^x=15 . Please continue.

    to #12:

    \log_7\left(\frac{x^4}{\sqrt[6]{y}}  \right)=\log_7\left(x^4 \cdot y^{-\frac16}  \right)=4\cdot \log_7(x)-\frac16 \cdot \log_7(y)

    to #13:

    use the property: \log(a) + \log(b) = \log(a \cdot b)

    to #14:

    use the property: \log(a) - \log(b) = \log\left(\frac ab\right)

    to #15:

    you are supposed to know: \log_b(a)=\frac{\ln(a)}{\ln(b)}=\frac{\log_{10}(a)  }{\log_{10}(b)}

    It doesn't matter which logarithmetic function you use. You should come out with
    \log_8(155) \approx 2.425374802...
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    I don't know that I really understand these...would you mind explaining what you did (like on number 9)? I'm really sorry, I'm having a tough time understanding this.
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    Quote Originally Posted by Raleigh View Post
    I don't know that I really understand these...would you mind explaining what you did (like on number 9)? I'm really sorry, I'm having a tough time understanding this.
    Hi,

    logarithm is only another word for exponent. The base to which this exponent belongs is added as an index. Thus \log_8() belongs to the base 8.

    If you use this base and the logarithm then the result of this power is the value or term in the bracket following the word "log":

    8^{\log_8 (4x+14)} = 4x+14
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  5. #5
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    Quote Originally Posted by Raleigh View Post
    I don't know that I really understand these...would you mind explaining what you did (like on number 9)? I'm really sorry, I'm having a tough time understanding this.
    Hello,

    I'll send you a .pdf-file with the most used formulas concerning powers and logarithms. Even though it is written in German you'll understand it easily.
    Attached Files Attached Files
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    I think that I may understand this now, but could someone please look over the work that I have? numbers 9 -15?
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    Quote Originally Posted by Raleigh View Post
    I think that I may understand this now, but could someone please look over the work that I have? numbers 9 -15?
    In problem 11 you have
    400 = 375(1 + e^{0.5x})

    Continuing:
    \frac{400}{375} = 1 + 3^{0.5x}
    (You left the 1 out on the RHS.)

    Problem 12: What's this under the answer with the 3x^0 etc. It is unrelated to your problem. (Which you got right, by the way. )

    Problem 15: The change of base formula is
    log_a(b) = \frac{log_c(b)}{log_c(a)}

    So
    log_8(155) = \frac{log_{10}(155)}{log_{10}(8)}

    This is not the same as
    log_{10} \left ( \frac{155}{8} \right )

    -Dan
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  8. #8
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    I appreciate the assistance very much! Thanks!
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  9. #9
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    Quote Originally Posted by Raleigh View Post
    I think that I may understand this now, but could someone please look over the work that I have? numbers 9 -15?
    Hi,

    to #8:

    you calculated \sin(\theta) and \csc(\theta)

    to #11:

    the result should be: x=\ln(15) \approx 2.708

    to #15:

    the result isn't written correctly: \frac{\log(155)}{\log(8)} \approx 2.425
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