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einsten_alkhawarizmi3@yahoo.com

Crux- $Natural Logarithm$

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$(x^2y-1)\ln x+(xy^2-1)\ln y=0$. Syaratnya $x,y\in\mathbb{R}^+$. Persamaannya ekuivalen dengan \begin{align*}\ln x^{x^2y-1}+\ln y^{xy^2-1}&=\ln 1\\ \ln (x^{x^2y-1}y^{xy^2-1})&=\ln 1\\ x^{x^2y-1}y^{xy^2-1}&=1.\end{align*} Maka diperoleh sistem persamaan $$\begin{cases}x^2y-1=0\implies xy=\frac{1}{x}\\xy^2-1=0\implies xy=\frac{1}{y}\end{cases}\implies x=y.$$ Dari persamaan 1, diperoleh $$x^3=1\implies x=1=y.$$ Jadi, solusinya adalah $(x,y)=\{(1,1)\}$.

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