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Mathematics 17 Online
OpenStudy (anonymous):

ln(x-2)<0 if and only if:

OpenStudy (anonymous):

Hmmm, since \(e^x\) is a one to one function, so I think you can use \(e^x\) on both sides.. \[ e^{\ln (x-2)} < e^0\\ x-2 < 1 \]

OpenStudy (anonymous):

Inequalities are a bit tricky because you have to be aware of things like sign changes.

OpenStudy (anonymous):

so x<3 ?

OpenStudy (anonymous):

Since \(e^x\) is always increasing, then this should maintain equality.

OpenStudy (anonymous):

Yes, but it might also be important to consider the domain of \(\ln(x)\) as well

OpenStudy (anonymous):

It not defined for non-positive numbers. That's why you need \(0<x-2\)

OpenStudy (anonymous):

aahh i see so its 2<x<3 correct?

OpenStudy (anonymous):

That looks right.

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