Question

log base 2 (4x)- log base 2(9-x)=1

Answer 1

$log_{2} (4x)- log_{2} (9-x)=1 \\ \\ log_{2} (4x)= 1+ log_{2} (9-x) \\ \\ log_{2} (4x)= log_{2}2 + log_{2} (9-x) \\ \\ log_{2} (4x)= log_{2} [2.(9-x)] \\ \\ log_{2} (4x)= log_{2} (18-2x) \\ \\ 4x=18-2x \\ \\ 6x=18 \\ \\ x= 3$