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Solve this equation: `ln(-2x)=lnx^2` for logarithms.
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High School Teacher
To eliminate the logarithm, apply the rule `e^(lnx)=x` .
Express the equation in quadratic form `ax^2+bx+c=0` .
Set each factor to zero and solve for x.
`x+2=0` and `x=0`
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` .
Posted by mjripalda on August 24, 2012 at 12:02 AM (Answer #1)
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