# arrow_back Solve the logarithmic equation $\ln (x-2)+\ln (2 x-3)=2 \ln x$

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Solve the logarithmic equation $\ln (x-2)+\ln (2 x-3)=2 \ln x$

(1) Use Power Rule: $\log _{b} x^{c}=c \log _{b} x$.
$\ln (x-2)+\ln (2 x-3)=\ln \left(x^{2}\right)$
(2) Use Product Rule: $\ln (x y)=\ln x+\ln y$.
$\ln ((x-2)(2 x-3))=\ln \left(x^{2}\right)$
3 Cancel "In" on both sides.
$(x-2)(2 x-3)=x^{2}$
(4) Expand.
$2 x^{2}-3 x-4 x+6=x^{2}$

5) Simplify $2 x^{2}-3 x-4 x+6$ to $2 x^{2}-7 x+6$.
$2 x^{2}-7 x+6=x^{2}$
(6) Move all terms to one side.
$2 x^{2}-7 x+6-x^{2}=0$
(7) Simplify $2 x^{2}-7 x+6-x^{2}$ to $x^{2}-7 x+6$.
$x^{2}-7 x+6=0$
8) Factor $x^{2}-7 x+6$.
$(x-6)(x-1)=0$

9 Solve for $x$.
$x=6,1$
10 Check solution
When $x=1$, the original equation $\ln (x-2)+\ln (2 x-3)=2 \ln x$ does not hold true. We will drop $x=1$ from the solution set.
11 Therefore,
$x=6$
by Bronze Status
(7,476 points)

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