Algebra question

**champrock** · February 16th 2009, 07:45 AM

hi

how to solve this?

thanks

**Soroban** · February 16th 2009, 10:22 AM

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)

**champrock** · February 16th 2009, 10:25 AM

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

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