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Thread: 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. $\displaystyle a^x = \left(a^x\right)^1$

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

    Since $\displaystyle b^y = a^x$ you get:

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

    According to the given conditions

    $\displaystyle a^x = \left(a^x\right)^1=\left(a^x\right)^{x+y}$ 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.

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

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


    Prove that if $\displaystyle a^x \:= \:b^y \:= \ab)^{xy}$, then: .$\displaystyle x+y \:= \:1$

    We have: .$\displaystyle a^x \:=\ab)^{xy}$

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


    We have: .$\displaystyle b^y \:=\ab)^{xy}$

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


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


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


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

    *

    If $\displaystyle x = 0$, we have: .$\displaystyle a^0 \:=\:b^y \:=\ab)^0\quad\Rightarrow\quad b^y \:=\:1$

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

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

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

    Hence: .$\displaystyle x \:\neq\:0$

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

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