Solve this equation: `ln(-2x)=lnx^2` for logarithms.

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`ln(-2x)=lnx^2`

To eliminate the logarithm, apply the rule `e^(lnx)=x` .

`e^(ln(-2x))=e^(lnx^2)`

`-2x=x^2`

Express the equation in quadratic form `ax^2+bx+c=0` .

`x^2+2x=0`

Then, factor.

`x(x+2)=0`

Set each factor to zero and solve for x.

`x+2=0` and `x=0`

`x=-2`

Then, substitute the values of x to the original equation to check.

`x=-2` , `ln (-2*-2) = ln(-2)^2`

`ln4 = ln4` (True)

`x=0 ` , `ln(-2*0)= ln(0)^2`

`ln 0 = ln 0` (Invalid logarithm)

Note that in logarithm, we cannot have a zero as the argument. It should always be greater than zero.

**Hence, the solution to the equation `ln(-2x)=lnx^2` is `x=-2` .**

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