What is x if #log_4 x= 1/2 + log_4 (x-1)#?

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Answer 1 · 2018-05-01T17:28:47

#x=2#

As #log_4 x= 1/2 + log_4 (x-1)#

#log_4x-log_4(x-1)=1/2#

or #log_4(x/(x-1))=1/2#

i.e. #x/(x-1)=4^(1/2)=2#

and #x=2x-2#

i.e. #x=2#

Answer 2 · 2018-05-01T17:36:01

# x=2#.

#log_4x=1/2+log_4(x-1)#.

#:. log_4 x-log_x(x-1)=1/2#.

#:. log_4{x/(x-1)}=1/2...[because, log_bm-log_bn=log_b(m/n)]#.

#:. {x/(x-1)}=4^(1/2)=2,...[because," the definition of "log]#.

#:. x=2(x-1)=2x-2#.

#:. -x=-2, or, x=2#.

This root satisfy the given eqn.

#:. x=2#.