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Question Number 87613 by mary_ last updated on 05/Apr/20

Commented by TANMAY PANACEA. last updated on 05/Apr/20

pls recheck question...

$${pls}\:{recheck}\:{question}... \\ $$

Answered by mind is power last updated on 05/Apr/20

(y−2x)ln(2)+ln(x+y)=ln(6.25)  (2x−y)ln(x+y)=ln(5)  ⇒(y−2x)ln(2).ln(x+y)=−ln(2)ln(5)  ⇒ln(x+y) and (y−2x)ln(2) are Solution of  X^2 −ln(6.25)X−ln(5)ln(2)=0  X∈{a,b)   { (((y−2x)ln(2)=a)),((ln(x+y)=b)) :}  ⇒x=((e^b −(a/(ln(2))))/3),y=e^b −((e^a −(a/(ln(2))))/3)  and x=((e^a −(b/(ln(2))))/3),y=e^a −(e^(b−(b/(ln(2)))) /3)

$$\left({y}−\mathrm{2}{x}\right){ln}\left(\mathrm{2}\right)+{ln}\left({x}+{y}\right)={ln}\left(\mathrm{6}.\mathrm{25}\right) \\ $$$$\left(\mathrm{2}{x}−{y}\right){ln}\left({x}+{y}\right)={ln}\left(\mathrm{5}\right) \\ $$$$\Rightarrow\left({y}−\mathrm{2}{x}\right){ln}\left(\mathrm{2}\right).{ln}\left({x}+{y}\right)=−{ln}\left(\mathrm{2}\right){ln}\left(\mathrm{5}\right) \\ $$$$\Rightarrow{ln}\left({x}+{y}\right)\:{and}\:\left({y}−\mathrm{2}{x}\right){ln}\left(\mathrm{2}\right)\:{are}\:{Solution}\:{of} \\ $$$${X}^{\mathrm{2}} −{ln}\left(\mathrm{6}.\mathrm{25}\right){X}−{ln}\left(\mathrm{5}\right){ln}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$${X}\in\left\{{a},{b}\right) \\ $$$$\begin{cases}{\left({y}−\mathrm{2}{x}\right){ln}\left(\mathrm{2}\right)={a}}\\{{ln}\left({x}+{y}\right)={b}}\end{cases} \\ $$$$\Rightarrow{x}=\frac{{e}^{{b}} −\frac{{a}}{{ln}\left(\mathrm{2}\right)}}{\mathrm{3}},{y}={e}^{{b}} −\frac{{e}^{{a}} −\frac{{a}}{{ln}\left(\mathrm{2}\right)}}{\mathrm{3}} \\ $$$${and}\:{x}=\frac{{e}^{{a}} −\frac{{b}}{{ln}\left(\mathrm{2}\right)}}{\mathrm{3}},{y}={e}^{{a}} −\frac{{e}^{{b}−\frac{{b}}{{ln}\left(\mathrm{2}\right)}} }{\mathrm{3}} \\ $$

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