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Math Help - logarithm bases assignment

  1. #1
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    logarithm bases assignment

    hello everyone,

    this is my first time on here, so i hope i'm doing it right

    we have a year 12 maths assignment due in about a week, and it says:

    Let log(a)x=c and log(b)x=d. Find the general statement that expresses log(ab)x in terms of c and d.

    is the answer log(ab)x=(cd)/(c+d)??

    if it is.. how on earth is it right? lol.. how can i prove it?

    thank you so much
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by Missy View Post
    hello everyone,

    this is my first time on here, so i hope i'm doing it right

    we have a year 12 maths assignment due in about a week, and it says:

    Let log(a)x=c and log(b)x=d. Find the general statement that expresses log(ab)x in terms of c and d.

    is the answer log(ab)x=(cd)/(c+d)??

    if it is.. how on earth is it right? lol.. how can i prove it?

    thank you so much

    Welcome to the Forum

    Formula



    <br />
log_N(M) = \frac{1}{log_{M}(N)}


    --------------------------------------------------

    log_x(a)+ log_x( b) = log_x(ab)

    <br />
\frac{1}{log_a(x)}+\frac{1}{log_b(x)} = \frac{1}{log_{ab}(x)}

    \frac{1}{c}+\frac{1}{d} =  \frac{1}{log_{ab}(x)}

    \frac{cd}{c+d} = log_{ab}(x)
    -------------------------------------------

    Was this a graded assignment ?
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  3. #3
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    ahh thank you thank you!
    umm.. it depends if its good enough.. my teacher will pick the best out of four assignments..
    thank you!!!
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  4. #4
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    Hello, Missy!

    Welcome aboard!

    Here's another approach . . . It's considerably longer,
    . . but it will impress/surprise/terrify your teacher.


    Let \log_aX \:=\:c and \log_bX\:=\:d
    Find the general statement that expresses \log_{ab}X in terms of c and d.

    is the answer: . \log_{ab}X\:=\:\frac{cd}{c+d} ?

    We have: . \begin{array}{ccccccccccc}\log_aX \:=\:c &&\Rightarrow&&  a^c \:=\:X &&\Rightarrow&& a \:=\:X^{\frac{1}{c}} & [1]\\ \log_bX \:=\:d && \Rightarrow& & b^d \:=\:X &&\Rightarrow&& b \:=\:X^{\frac{1}{d}} & [2] \end{array}


    Multiply [1] and [2]: . ab \:=\:X^\frac{1}{c}\cdot X^{\frac{1}{d}} \;=\;X^{\frac{1}{c} + \frac{1}{d}} \quad\Rightarrow\quad ab\;=\;X^{\frac{c+d}{cd}}


    Raise both sides to the power \frac{cd}{c+d}

    . . (ab)^{\frac{cd}{c+d}} \;=\;\left(X^{\frac{c+d}{cd}}\right)^{\frac{cd}{c+  d}} \quad\Rightarrow\quad (ab)^{\frac{cd}{c+d}} \;=\;X


    Take logs, base ab

    . . \log_{ab}\left[(ab)^{\frac{cd}{c+d}}\right] \;=\;\log_{ab}(X)  \quad\Rightarrow\quad \log_{ab}(X) \;=\;\frac{cd}{c+d}\cdot\underbrace{\log_{ab}(ab)}  _{\text{This is 1}}

    Therefore: . \log_{ab}(X) \:=\:\frac{cd}{c+d}

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  5. #5
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    Therefore: . \log_{ab}(X) \:=\:\frac{cd}{c+d}

    [/quote]
    i agree with u, but the thing is, i dont understand how u can substitute values for a, b, and x .(for testing the formula)
    soo can someone plz help me, cuz i am having this problem.
    and can u do the subtitution on the same formula!!!! :P


    . \log_{ab}(X) \:=\:\frac{cd}{c+d}

    thnx

    plzzz help me i need this for tomorrow
    Last edited by jack8; December 21st 2009 at 03:20 AM.
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