Question

log2 (3x+7)=4 rezolvare a

Answer 1

$log_{2}(3x+7)=4,DVA:x\ \textgreater \ - \frac{7}{3}$
$3x+7=16$
$3x=9$
$x=3$
S={3}

Answer 2

Conditia pentru existenta:
$3x+7\ \textgreater \ 0\\x+(7/3)\ \textgreater \ 0\\x\ \textgreater \ 0-(7/3)\\x\ \textgreater \ -\frac{7}{3}\\Rezolvare:\\Log2(3x+7)=4\\3x+7=2^{4}\\3x+7=2*2*2*2\\ 3x+7=16\\3x=16-7\\3x=9\\x=\frac{9}{3}\\x=3$
S={3}