1. ## Algebra question

hi

how to solve this?

thanks

2. Hello, champrock!

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

The relation between $\displaystyle x$ and $\displaystyle y$ is:

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

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

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

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

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

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

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