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

Jun 24, 2016

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

Explanation:

${\log}_{6} \left(2 x - 1\right) + {\log}_{6} x = 2$

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

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

Now splitting the middle term

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

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

$\left(x + 4\right) \left(2 x - 9\right) = 0$

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

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

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