How do you solve $log_6 (2x-1) + log_6 x = 2$ ?

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Answer 1 · 2016-06-24T06:30:22

Solution is $x=9/2$

$log_6(2x-1)+log_6x=2$

$hArrlog_6x(2x-1)=2log_6(6)=log_6(36)$

Hence $x(2x-1)=36$ or $2x^2-x-36=0$

Now splitting the middle term

$2x^2-9x+8x-36=0$ or

$x(2x-9)+4(2x-9)=0$ or

$(x+4)(2x-9)=0$

i.e. either $x+4=0$ i.e. $x=-4$ (but this is out of domain)

or $2x-9=0$ i.e. $x=9/2$

Hence, solution is $x=9/2$

How do you solve log_6 (2x-1) + log_6 x = 2log_6 (2x-1) + log_6 x = 2?