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.

$:. x=2$ .

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