Question

log2(3x-3)=4 .................

Answer 1

Найдём одз:

3x-3>0

3x>3

x>1

$3x-3=2^4$

$3x-3=16$

$3x=19$

$x=\frac{19}{3}=6\frac{1}{3}$

Answer 2

log2(3x-3)=4 => x>1

3x-3=2^4 =>

3x = 19 =>

x=19/3 или x=6 целых 1/3