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  1. #1
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    Gg

     <br />
4^\times=27<br />

     <br />
9^y=8<br />

    x.y=?
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by OPETH View Post
     <br />
4^\times=27<br />

     <br />
9^y=8<br />

    x.y=?
    4^x=27

    \log 4^x=\log27

    x\log4=\log 27

    x=\frac{\log27}{\log4}

    x\approx2.3774
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by OPETH View Post
     <br />
4^\times=27<br />

     <br />
9^y=8<br />

    x.y=?
    This can also be solved without using logarithms :

    \begin{cases}4^x=27 \\ 9^y=8 \end{cases} \Longleftrightarrow \begin{cases}2^{2x}=3^3 & (1) \\ 3^{2y}=2^3 & (2)\end{cases}

    From (2) we get that 2=3^{\frac{2y}{3}} which gives, when substituting in (1) :

    \left(3^{\frac{2y}{3}}\right)^{2x}=3^3 \Longleftrightarrow 3^{\frac{4xy}{3}}=3^3

    hence xy=\ldots
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by masters View Post
    4^x=27

    \log 4^x=\log27

    x\log4=\log 27

    x=\frac{\log27}{\log4}

    x\approx2.3774
    I obviously misread the exercise, but I'll finish up the mess that I started to hopefully arrive at the same conclusion as flyingsquirrel.

    9^y=8

    \log 9^y=\log8

    y\log9=\log 8

    y=\frac{\log8}{\log9}

    xy=\frac{\log27}{\log4}\cdot\frac{\log8}{log9}=\bo  xed{2.25}
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  5. #5
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    Hello, OPETH!

    Your answer is correct, masters!

    Here's a slight variation of flyingsquirrel's solution . . .


    Given: . \begin{array}{ccc}4^x &=& 27 \\ 9^y &=& 8 \end{array}

    . . . . . . xy\:=\:?

    We have: . 4^x\:=\:27\quad\Rightarrow\quad(2^2)^x \:=\:27 \quad\Rightarrow\quad 2^{2x} \:=\:3^3 \quad\Rightarrow\quad 2 \:=\:3^{\frac{3}{2x}}\;\;{\color{blue}[1]}

    . . . . . . . . 9^y \:=\:8 \quad\Rightarrow\quad (3^2)^y \:=\:8 \quad\Rightarrow\quad 3^{2y} \:=\:2^3 \quad\Rightarrow\quad 2 \;=\:3^{\frac{2y}{3}}\;\;{\color{blue}[2]}


    Equate [1] and [2]: . 3^{\frac{3}{2x}}\:=\:3^{\frac{2y}{3}} \quad\Rightarrow\quad \frac{3}{2x}\:=\:\frac{2y}{3}\quad\Rightarrow\quad 4xy \:=\:9

    Therefore: . \boxed{xy \:=\:\frac{9}{4}}

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