# How do you solve ln(2x-3)+ln(x-2) = 2lnx?

Jun 15, 2016

Solution is $x = 6$

#### Explanation:

$\ln \left(2 x - 3\right) + \in \left(x - 2\right) = 2 \ln x$

or $\ln \left(2 x - 3\right) \left(x - 2\right) = \ln {x}^{2}$

or $\left(2 x - 3\right) \left(x - 2\right) = {x}^{2}$

or $2 {x}^{2} - 4 x - 3 x + 6 - {x}^{2} = 0$

or ${x}^{2} - 7 x + 6 = 0$

or ${x}^{2} - 6 x - x + 6 = 0$

or $x \left(x - 6\right) - 1 \left(x - 6\right) = 0$

or $\left(x - 1\right) \left(x - 6\right) = 0$

i.e. $x = 1$ or $x = 6$

But $x = 1$ does not lie in domain as this makes $2 x - 3$ and $x - 2$ both negative, hence

Solution is $x = 6$

Jun 15, 2016

$\therefore x = 6.$

#### Explanation:

$\ln \left(2 x - 3\right) + \ln \left(x - 2\right) = 2 \ln x = \ln {x}^{2} ,$... [since $m \ln a = \ln {a}^{m}$]
$\therefore \ln \left\{\left(2 x - 3\right) \left(x - 2\right)\right\} = \ln {x}^{2} ,$... [as $\ln a + \ln b = \ln \left(a b\right)$]
$\therefore \ln \left(2 {x}^{2} - 3 x - 4 x + 6\right) = \ln {x}^{2.}$
$\therefore \ln \left(2 {x}^{2} - 7 x + 6\right) = \ln {x}^{2.}$
$\therefore 2 {x}^{2} - 7 x + 6 = {x}^{2.}$...[because $\ln$ is $1 \to 1$ fun.]
$\therefore {x}^{2} - 7 x + 6 = 0.$
$\therefore \left(x - 6\right) \left(x - 1\right) = 0.$
$\therefore x = 6 , x = 1.$

But $x = 1$ is not permissible, as it makes both $\ln \left(x - 2\right)$ & $\ln \left(2 x - 3\right) \left[= \ln \left(- 1\right)\right]$ undefined.$\therefore x \ne 1.$

We verify $x = 6$ : $L H S = \ln \left(12 - 3\right) + \ln \left(6 - 2\right) = \ln 9 + \ln 4 = \ln \left(4 \cdot 9\right) = \ln 36 = \ln {6}^{2} = 2 \ln 6 = R H S .$

$\therefore x = 6 ,$ is the only soln.