2log_9(2(1/2)^x-1)=log_27((1/4)^x-4)^3
=>2xxlog_9(27)log_27(2(1/2)^x-1)=log_27((1/4)^x-4)^3
=>2xxlog_9(9^(3/2))log_27(2(1/2)^x-1)=log_27((1/4)^x-4)^3
=>2xx3/2log_9(9)log_27(2(1/2)^x-1)=log_27((1/4)^x-4)^3
=>3log_27(2(1/2)^x-1)=log_27((1/4)^x-4)^3
=>log_27(2(1/2)^x-1)^3=log_27((1/4)^x-4)^3
=>(2(1/2)^x-1)=(((1/2)^x)^2-4)
Taking (1/2)^x=y we get
(2y-1)=(y^2-4)
=>y^2-2y-3=0
=(y+1)(y-3)=0
So (1/2)^x=-1->"not possible"
And (1/2)^x=3
=>-xlog2=log3
=>x=-log3/log2