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Math Help - Logarithm help!

  1. #1
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    Logarithm help!

    Hey guys, I appreciate the help on these logs and I ran into another snag

    I tried to use logarit for these but I keep getting the wrong answer

    1. 3log(base 2) x + 1/2 log(base 2) x - 3log (base 2) x - 5

    2. If log (base b) 3 = 0.56 and log (base b) 2 = 0.36 evaluate...

    log (base b) square root of 2/3

    3. Solve the equation for x. Give an exact answer and a four place decimal approximation

    log (2x+1) = -0.9

    4. Solve

    log (base 2) x + log (base 2) (3x-2) = 3

    PLEASE PLEASE PLEASE HELP!!! Answers are greatly appreciated!
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  2. #2
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    Hello Croyt!

    Quote Originally Posted by Croyt View Post
    1. 3\log_2(x) + \frac12\log_2(x) - 3\log_2(x) - 5
    Quote Originally Posted by Croyt View Post
    2. If \log_b(3) = 0.56 and \log_b(2) = 0.36 evaluate \log_b\sqrt{\frac23}
    Here are some basic rules that will help you:

    1. a\log x=\log x^a

    2. \log xy=\log x + \log y

    3. \log\frac xy=\log x - \log y

    Quote Originally Posted by Croyt View Post
    3. Solve the equation for x. Give an exact answer and a four place decimal approximation

    \log(2x+1) = -0.9
    Are you dealing with base \text e, or base 10? In either case, the inverse operation of the logarithm is exponentiation. So, for example,

    \ln(2x+1)=-0.9

    \Rightarrow \text e^{\ln(2x+1)}=\text e^{-0.9}

    \Rightarrow 2x+1=\text e^{-0.9}

    Quote Originally Posted by Croyt View Post
    4. Solve

    \log_2(x) + \log_2(3x-2) = 3
    Use the properties I listed above to simplify, and then solve it like number 3.
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