Jump to content

Crux- $Natural Logarithm$

Recommended Posts

$(x^2y-1)\ln x+(xy^2-1)\ln y=0$. 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 $(x,y)=\{(1,1)\}$.

Share this post

Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now