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Math Help - Logarithms...again

  1. #1
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    Logarithms...again

    log(base 2) (x+4) + log(base 2) (x+8)=5


    Here's my work:
    log(2) (x+4)(x+8)=10
    x^2+12x+22=0
    (x+11) (x+2)
    -11,-2-->Answers

    But my teacher said you don't use negative numbers, so what's my answer?


    And, I don't know how to do this problem:

    log(base 5)(2x+4) - log(base 5)(x-1)=1

    My work:
    log5 2(x+2)/(x-1)=5

    Then what?
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  2. #2
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    Krizalid's Avatar
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    Quote Originally Posted by beetz View Post
    log(base 2) (x+4) + log(base 2) (x+8)=5
    \log_2(x+4)+\log_2(x+8)=5\implies\log_2\Big[(x+4)(x+8)\Big]=5.

    So, (x+4)(x+8)=32\implies x^2+12x=0\,\therefore\,x_1=0\,\wedge\,x_2=-12.

    We require that x+4>0 & x+8>0 such that the logarithms be well defined.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    \log_2(x+4)+\log_2(x+8)=5\implies\log_2\Big[(x+4)(x+8)\Big]=5.

    So, (x+4)(x+8)=32\implies x^2+12x=0\,\therefore\,x_1=0\,\wedge\,x_2=-12.

    We require that x+4>0 & x+8>0 such that the logarithms be well defined.
    my teacher said what you set the equation equal to is the base, in this case 2, multiplied by the answer given, 5.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by beetz View Post
    my teacher said what you set the equation equal to is the base, in this case 2, multiplied by the answer given, 5.
    no. if \log_a b = c then a^c = b

    in other words, what you set a log equal to is the power of the base (that is, the power you must raise the base to to get what's being logged). (you should pay attention in class)
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  5. #5
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    I was paying attention and I promise you that is exactly what he said. I even have a problem he did on the board and he multiplied the log base and the answer given. In fact, he never did a formal lecture, just skimmed over the homework and did a few problems.

    I think I get it...
    Thanks.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by beetz View Post
    I was paying attention and I promise you that is exactly what he said. I even have a problem he did on the board and he multiplied the log base and the answer given. In fact, he never did a formal lecture, just skimmed over the homework and did a few problems.
    ok, ok, if you say so. i believe you


    I think I get it...
    Thanks.
    show us the solution
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