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Math Help - Logarithms

  1. #1
    Member smmmc's Avatar
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    Logarithms

    Hey need help on the 3 questions,

    thanks heaps
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  2. #2
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    y = 3\log_{2}(5x) + 2

    You have to know that for your basic logarithm function y = \log_{b}x the asymptote is at x = 0. The only way that it would change is if we have a horizontal shift. Do we in y = 3\log_{2}(5x) + 2? (In other words what do the 3, 5, and +2 do?) If you can answer that, then you have the answer to the problem.


    As for #9,
    Which one of the following functions has a graph with a veritcal asymptote with equation x = b?
    Here's a hint: two of the choices can be eliminated because they do not have vertical asymptotes at all (instead, they only have a horizontal asymptote). Do you know which 2?


    01
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  3. #3
    Member smmmc's Avatar
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    its a toss up between a and b hrmm
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  4. #4
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    Hello, smmmc!

    I have a solution for #8, but there must be a better way.


    Given 3^x \:=\:4^y \:=\:12^z, show that: . z \:=\:\frac{xy}{x+y}

    We have: . 3^x \:=\:12^z

    . . Take logs: . \log(3^x) \:=\:\log(12^z) \quad\Rightarrow\quad x\log3 \:=\:z\log12

    . . And we have: . \log3 \:=\:\frac{z\log12}{x} .[1]


    We have: . 4^y \:=\:12^z

    . . Take logs: . \log(4^y) \:=\:\log(12^z) \quad\Rightarrow\quad y\log4 \:=\:z\log12

    . . And we have: . \log4 \:=\:\frac{z\log12}{y} .[2]


    \text{Add }{\color{blue}[1]}\text{ and }{\color{blue}[2]}\!:\quad \underbrace{\log3 + \log 4}_{\text{This is }\log12} \;=\;\frac{z\log12}{x} + \frac{z\log12}{y} \;=\;z\log12\left(\frac{x+y}{xy}\right)

    Hence: . \log12 \;=\;z\log12\left(\frac{x+y}{xy}\right) \quad\Rightarrow\quad 1 \;=\;z\left(\frac{x+y}{xy}\right)

    . . Therefore: . z \;=\;\frac{xy}{x+y}

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