How do you solve ${log}_{6} (2 x - 1) + {log}_{6} x = 2$ $log_6 (2x-1) + log_6 x = 2$ ?

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Answer 1 · 2016-06-24T06:30:22

Solution is $x = \frac{9}{2}$ $x=9/2$

${log}_{6} (2 x - 1) + {log}_{6} x = 2$ $log_6(2x-1)+log_6x=2$

$\Leftrightarrow {log}_{6} x (2 x - 1) = 2 {log}_{6} (6) = {log}_{6} (36)$ $hArrlog_6x(2x-1)=2log_6(6)=log_6(36)$

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

Now splitting the middle term

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

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

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

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

or $2 x - 9 = 0$ $2x-9=0$ i.e. $x = \frac{9}{2}$ $x=9/2$

Hence, solution is $x = \frac{9}{2}$ $x=9/2$

How do you solve log_6 (2x-1) + log_6 x = 2log6(2x−1)+log6x=2log_6 (2x-1) + log_6 x = 2?