Question Number 87613 by mary_ last updated on 05/Apr/20
Commented by TANMAY PANACEA. last updated on 05/Apr/20
$${pls}\:{recheck}\:{question}… \\ $$
Answered by mind is power last updated on 05/Apr/20
$$\left({y}−\mathrm{2}{x}\right){ln}\left(\mathrm{2}\right)+{ln}\left({x}+{y}\right)={ln}\left(\mathrm{6}.\mathrm{25}\right) \\ $$$$\left(\mathrm{2}{x}−{y}\right){ln}\left({x}+{y}\right)={ln}\left(\mathrm{5}\right) \\ $$$$\Rightarrow\left({y}−\mathrm{2}{x}\right){ln}\left(\mathrm{2}\right).{ln}\left({x}+{y}\right)=−{ln}\left(\mathrm{2}\right){ln}\left(\mathrm{5}\right) \\ $$$$\Rightarrow{ln}\left({x}+{y}\right)\:{and}\:\left({y}−\mathrm{2}{x}\right){ln}\left(\mathrm{2}\right)\:{are}\:{Solution}\:{of} \\ $$$${X}^{\mathrm{2}} −{ln}\left(\mathrm{6}.\mathrm{25}\right){X}−{ln}\left(\mathrm{5}\right){ln}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$${X}\in\left\{{a},{b}\right) \\ $$$$\begin{cases}{\left({y}−\mathrm{2}{x}\right){ln}\left(\mathrm{2}\right)={a}}\\{{ln}\left({x}+{y}\right)={b}}\end{cases} \\ $$$$\Rightarrow{x}=\frac{{e}^{{b}} −\frac{{a}}{{ln}\left(\mathrm{2}\right)}}{\mathrm{3}},{y}={e}^{{b}} −\frac{{e}^{{a}} −\frac{{a}}{{ln}\left(\mathrm{2}\right)}}{\mathrm{3}} \\ $$$${and}\:{x}=\frac{{e}^{{a}} −\frac{{b}}{{ln}\left(\mathrm{2}\right)}}{\mathrm{3}},{y}={e}^{{a}} −\frac{{e}^{{b}−\frac{{b}}{{ln}\left(\mathrm{2}\right)}} }{\mathrm{3}} \\ $$