2log32-log3(x-1)=1+log35 log3 основание

Question

2log32-log3(x-1)=1+log35

log3 основание

Answer 1

lg3(x-1)=lg3(4)-1-lg3(5)

lg3(x-1)=lg3(4/15)

x-1=4/15

x=1+4/15=19/15