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Math Help - Another problem in exponential and logarithms

  1. #1
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    Another problem in exponential and logarithms

    Prove that if a^x = b^y = (ab)^xy, then x+y = 1.

    Please show me how to do this.
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  2. #2
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    Quote Originally Posted by geton View Post
    Prove that if a^x = b^y = (ab)^xy, then x+y = 1.

    Please show me how to do this.
    Hello,

    1. a^x = \left(a^x\right)^1

    2. (ab)^{x \cdot y} = \left( (ab)^x \right)^y = a^{xy} \cdot b^{xy} = a^{xy} \cdot b^{yx}

    Since b^y = a^x you get:

    a^{xy} \cdot a^{x \cdot x} = \left(a^x\right)^{x+y}

    According to the given conditions

    a^x = \left(a^x\right)^1=\left(a^x\right)^{x+y} and therefore x + y = 1
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  3. #3
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    I believe there are a few ways to do this.

    a^x = b^y = (ab)^{xy}
    \implies \log a^x = \log b^y = \log (ab)^{xy}
    \implies x\log a = y\log b = (xy)\log (ab)
    \implies x\log a = y\log b = (xy)\log a + (xy)\log b
    \implies y(y\log b) + (xy)\log b = y\log b
    \implies y^2\log b + (xy)\log b = y\log b
    \implies y^2 + xy = y
    \implies y+x = 1
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  4. #4
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    Hello, geton!

    Sean is right . . . There must be dozens of approaches.


    Prove that if ab)^{xy}" alt="a^x \:= \:b^y \:= \ab)^{xy}" />, then: .  x+y \:= \:1

    We have: . ab)^{xy}" alt="a^x \:=\ab)^{xy}" />

    Take logs: . x\log(a) \:=\:xy\log(ab)
    * Divide by x\!:\;\;\log(a) \:=\:y\log(ab)\quad\Rightarrow\quad y \:=\:\frac{\log(a)}{\log(ab)} .[1]


    We have: . ab)^{xy}" alt="b^y \:=\ab)^{xy}" />

    Take logs: . y\log(b) \:=\:xy\log(ab)
    * Divide by y\!:\;\;\log(b) \:=\:x\log(ab)\quad\Rightarrow\quad x \:=\:\frac{\log(b)}{\log(ab)} .[2]


    Add [1] and [2]: . x + y \;=\;\frac{\log(a)}{\log(ab)} + \frac{\log(b)}{\log(ab)} \;=\;\frac{\log(a)+\log(b)}{\log(ab)} \;=\;\frac{\log(ab)}{\log(ab)}


    . . Therefore: . \boxed{x + y \:=\:1}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    *

    If x = 0, we have: . ab)^0\quad\Rightarrow\quad b^y \:=\:1" alt="a^0 \:=\:b^y \:=\ab)^0\quad\Rightarrow\quad b^y \:=\:1" />

    There are two possibilities: . \begin{array}{cc}{\color{red}(1)} &y \:= \:0 \\ {\color{red}(2)} & b \:= \:1 \end{array}

    In case (1), x + y \:\neq \:1

    In case (2), b is strictly determined which contradicts the original equation
    . . which is true for all values of a\text{ and }b.

    Hence: . x \:\neq\:0

    Similarly, we can show that: . y \:\neq \:0

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