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Math Help - Algebra question

  1. #1
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    Algebra question

    hi

    how to solve this?



    thanks
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  2. #2
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    Hello, champrock!

    \text{If }\:\begin{Bmatrix}x &=& t^{\frac{1}{t-1}} & {\color{blue}[1]} \\ y &=& t^{\frac{t}{t-1}} & {\color{blue}[2]} \end{Bmatrix}\quad t > 0,\:t \neq 1

    The relation between x and y is:

    . . (a)\;y^x \:=\:x^{\frac{1}{x}} \qquad (b)\;x^{\frac{1}{y}} \:=\:y^{\frac{1}{x}} \qquad (c)\;x^y \:=\:y^x\qquad (d)\;x^y \:=\:y^{\frac{1}{x}}

    From [2], we have: . y \;=\;t^{\frac{t}{t-1}} \;=\;\left(t^{\frac{1}{t-1}}\right)^t \quad\Rightarrow\quad y \:=\:x^t

    Take logs: . \ln(y) \:=\:\ln\left(x^t\right) \quad\Rightarrow\quad \ln(y) \:=\:t\cdot\ln(x) \quad\Rightarrow\quad t \:=\:\frac{\ln y}{\ln x} .[3]


    Divide [2] by [1]: . \frac{y}{x} \;=\;\frac{t^{\frac{t}{t-1}} }{t^{\frac{1}{t-1}}} \quad\Rightarrow\quad \frac{y}{x} \;=\;t .[4]


    Equate[4] and [3]: . \frac{y}{x} \:=\:\frac{\ln(y)}{\ln(x)} \quad\Rightarrow\quad y\!\cdot\!\ln(x) \:=\:x\!\cdot\!\ln(y) \quad\Rightarrow\quad \ln\left(x^y\right) \:=\:\ln\left(y^x\right)


    Therefore: . x^y \:=\:y^x . . . answer (c)

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  3. #3
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    wow! thanks a ton fr that great and simple solution.
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