The value of x satisfying #2log_9(2(1/2)^x-1)=log_27((1/4)^x-4)^3# is?

1 Answer
Aug 1, 2018

#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#