kolutastmr

2022-07-02

Logarithmic system of equations

fugprurgeil

Notice
$\mathrm{ln}x=2\mathrm{ln}y\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}x=\mathrm{ln}{y}^{2}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x={y}^{2}$
and
${3}^{x}=27y\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{3}^{x}={3}^{3}y\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{3}^{x-3}=y\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{3}^{{y}^{2}-3}=y\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}3\left({y}^{2}-3\right)=y\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{y}^{2}-3-\frac{y}{\mathrm{ln}3}=0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{y}^{2}-\frac{1}{\mathrm{ln}3}y-3=0$
This is a quadratic equation, which you can easily solve.

Palmosigx

The first equation says that $x,y>0$ and then it's equivalent to $x={e}^{\mathrm{ln}x}={e}^{3\mathrm{ln}y}={e}^{\mathrm{ln}{y}^{3}}={y}^{3}$.
The second equation is equivalent to $x=3y$ by taking a logarithm, so we have
$3y={y}^{3}⟺y\left(y-\sqrt{3}\right)\left(y+\sqrt{3}\right)=0$
and the only solution is $y=\sqrt{3}$, which gives $x=3\sqrt{3}$.

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