Question

Va Rog ajutati-ma repede

Answer 1

 

$\displaystyle\bf\\ m_g~a~numerelor~~3^x~si~9^x=9\\\\m_g=\sqrt{3^x\cdot9^y}=9\\\\ \text{\bf Se stie ca }~x=y+1\\\\\sqrt{3^{y+1}\cdot9^y}=9\\\\\sqrt{3^{y+1}\cdot\Big(3^2\Big)^y}=9\\\\\sqrt{3^{y+1}\cdot3^{2y}}=9\\\\\sqrt{3^{y+1+2y}}=3^2\\\\\sqrt{3^{3y+1}}=3^2~~\Big|~ridicam~la~puterea~a~2-a\\\\\Big(\sqrt{3^{3y+1}}\Big)^2=\Big(3^2\Big)^2\\\\3^{3y+1}=3^{2\cdot2}\\\\3^{3y+1}=3^{4}\\\\3y+1=4\\\\3y=4-1\\\\3y=3\\\\y=\frac{3}{3}\\\\\boxed{\bf~y=1}\\\\x=y+1\\\\x=1+1\\\\\boxed{\bf~x=2}$