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

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smullen12 | Student, Grade 11 | eNoter

Posted August 23, 2012 at 11:21 PM via web

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

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Mary Joy Ripalda | High School Teacher | (Level 3) Educator

Posted August 24, 2012 at 12:02 AM (Answer #1)

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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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