Question

Помогите с логарифмами.
$\left \{ {{ log _{3}x+log _{3}y=1-log _{3}2} \atop {log _{3}(x+y)=2}} \right.$

Answer 1

$\displaystyle\mathtt{\left\{{{\log_3x+\log_3y+\log_32=1}\atop{\log_3(x+y)=2}}\right\left\{{{\left\{{{\log_3(2xy)=1}\atop{x+y=9}}\right}\atop{\left\{{{x\ \textgreater \ 0}\atop{y\ \textgreater \ 0}}\right}}\right\left\{{{\left\{{{2xy=3}\atop{x=9-y}}\right}\atop{\left\{{{x\ \textgreater \ 0}\atop{y\ \textgreater \ 0}}\right}}\right}$

$\displaystyle\mathtt{\left\{{{\left\{{{2(9-y)y=3}\atop{x=9-y}}\right}\atop{\left\{{{x\ \textgreater \ 0}\atop{y\ \textgreater \ 0}}\right}}\right\left\{{{\left\{{{2y^2-18y+3=0}\atop{x=9-y}}\right}\atop{\left\{{{x\ \textgreater \ 0}\atop{y\ \textgreater \ 0}}\right}}\right\left\{{{\left\{{{\left[\begin{array}{ccc}\mathtt{y_1=\frac{9-5\sqrt{3}}{2}}\\\mathtt{y_2=\frac{9+5\sqrt{3}}{2}}\end{array}\right}\atop{x=9-y}}\right}\atop{\left\{{{x\ \textgreater \ 0}\atop{y\ \textgreater \ 0}}\right}}\right}$

ответ: $\mathtt{(\frac{9+5\sqrt{3}}{2};~\frac{9-5\sqrt{3}}{2})}$ или $\mathtt{(\frac{9-5\sqrt{3}}{2};~\frac{9+5\sqrt{3}}{2})}$