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Question Number 53740 by F_Nongue last updated on 25/Jan/19

please help me to solve this sistem: { ((log_2 (x+2y)−log_3 (x−2y)=2)),((x^2 −4y^2 =4)) :}

$${please}\:{help}\:{me}\:{to}\:{solve}\:{this}\: \\ $$$${sistem}: \\ $$$$\begin{cases}{{log}_{\mathrm{2}} \left({x}+\mathrm{2}{y}\right)−{log}_{\mathrm{3}} \left({x}−\mathrm{2}{y}\right)=\mathrm{2}}\\{{x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} =\mathrm{4}}\end{cases} \\ $$

Answered by Smail last updated on 25/Jan/19

x^2 −4y^2 =4⇒x=+_− (√(4+4y^2 ))=+_− 2(√(1+y^2 )) log_2 (x+2y)−log_3 (x−2y)=2 ((ln(x+2y))/(ln2))−((ln(x−2y))/(ln3))=2 ((ln(2(√(1+y^2 ))+2y))/(ln2))−((ln(2(√(1+y^2 ))−2y))/(ln3))=2 1+((ln((√(1+y^2 ))+y))/(ln2))−((ln2)/(ln3))−((ln((√(1+y^2 ))−y))/(ln3))=2 1+((ln((√(1+y^2 ))+y))/(ln2))−((ln2)/(ln3))−((ln(((1+y^2 −y^2 )/((√(1+y^2 ))+y))))/(ln3))=2 1+((sinh^(−1) (y))/(ln2))−log_3 2+((sinh^(−1) (y))/(ln3))=2 sinh^(−1) (y)(((ln3+ln2)/(ln2×ln3)))=1+((ln2)/(ln3))=((ln3+ln2)/(ln3)) sinh^(−1) (y)=ln2 y=sinh(ln2)=((e^(ln2) −e^(−ln2) )/2) y=(3/4) x=2(√(1+y^2 ))=2(√(1+(9/(16)))) x=(5/2)

$${x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} =\mathrm{4}\Rightarrow{x}=\underset{−} {+}\sqrt{\mathrm{4}+\mathrm{4}{y}^{\mathrm{2}} }=\underset{−} {+}\mathrm{2}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} } \\ $$$${log}_{\mathrm{2}} \left({x}+\mathrm{2}{y}\right)−{log}_{\mathrm{3}} \left({x}−\mathrm{2}{y}\right)=\mathrm{2} \\ $$$$\frac{{ln}\left({x}+\mathrm{2}{y}\right)}{{ln}\mathrm{2}}−\frac{{ln}\left({x}−\mathrm{2}{y}\right)}{{ln}\mathrm{3}}=\mathrm{2} \\ $$$$\frac{{ln}\left(\mathrm{2}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }+\mathrm{2}{y}\right)}{{ln}\mathrm{2}}−\frac{{ln}\left(\mathrm{2}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }−\mathrm{2}{y}\right)}{{ln}\mathrm{3}}=\mathrm{2} \\ $$$$\mathrm{1}+\frac{{ln}\left(\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }+{y}\right)}{{ln}\mathrm{2}}−\frac{{ln}\mathrm{2}}{{ln}\mathrm{3}}−\frac{{ln}\left(\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }−{y}\right)}{{ln}\mathrm{3}}=\mathrm{2} \\ $$$$\mathrm{1}+\frac{{ln}\left(\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }+{y}\right)}{{ln}\mathrm{2}}−\frac{{ln}\mathrm{2}}{{ln}\mathrm{3}}−\frac{{ln}\left(\frac{\mathrm{1}+{y}^{\mathrm{2}} −{y}^{\mathrm{2}} }{\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }+{y}}\right)}{{ln}\mathrm{3}}=\mathrm{2} \\ $$$$\mathrm{1}+\frac{{sinh}^{−\mathrm{1}} \left({y}\right)}{{ln}\mathrm{2}}−{log}_{\mathrm{3}} \mathrm{2}+\frac{{sinh}^{−\mathrm{1}} \left({y}\right)}{{ln}\mathrm{3}}=\mathrm{2} \\ $$$${sinh}^{−\mathrm{1}} \left({y}\right)\left(\frac{{ln}\mathrm{3}+{ln}\mathrm{2}}{{ln}\mathrm{2}×{ln}\mathrm{3}}\right)=\mathrm{1}+\frac{{ln}\mathrm{2}}{{ln}\mathrm{3}}=\frac{{ln}\mathrm{3}+{ln}\mathrm{2}}{{ln}\mathrm{3}} \\ $$$${sinh}^{−\mathrm{1}} \left({y}\right)={ln}\mathrm{2} \\ $$$${y}={sinh}\left({ln}\mathrm{2}\right)=\frac{{e}^{{ln}\mathrm{2}} −{e}^{−{ln}\mathrm{2}} }{\mathrm{2}} \\ $$$${y}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${x}=\mathrm{2}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{1}+\frac{\mathrm{9}}{\mathrm{16}}} \\ $$$${x}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$

Commented by F_Nongue last updated on 27/Jan/19

please sir can you explain your steps?

$${please}\:{sir}\:{can}\:{you}\:{explain}\:{your}\:{steps}? \\ $$

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