Question

Rezolvati in multimea numerelor reale ecuatia: 9^[(x+2)/2] + 3^(x+1) = 36. Multumesc

Answer 1

$9^{ \frac{x+2}{2} } + 3^{x+1} =36$
$3^{2* \frac{x+2}{2} } + 3^{x+1} =2 ^{2} *3 ^{2}$
$3^{x+2} +3 ^{x+1} =2 ^{2} *3 ^{2}$
$3 ^{x+1} *(3+1)=2 ^{2} *3 ^{2}$
$3^{x+1} *2 ^{2} =2 ^{2} *3 ^{2} =\ \textgreater \ x+1=2 =\ \textgreater \ x=1$