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Mathematics 8 Online
OpenStudy (chillhill):

find the domain of f(x)= √(ln(x-2))

OpenStudy (anonymous):

log (x-2 ) must be grater then zero so x is

OpenStudy (misty1212):

HI!!

OpenStudy (misty1212):

actually you need to solve \[\ln(x-2)\geq 0\]

OpenStudy (misty1212):

do you know how to do that?

OpenStudy (chillhill):

no

OpenStudy (er.mohd.amir):

log 1=0

OpenStudy (er.mohd.amir):

so x=2 is one solution?

OpenStudy (anonymous):

\((x-2)> 0\)

OpenStudy (anonymous):

In fact, f(x)=lnx ln(x)≥0 --> e^ln(x)≥e^0 --> x≥1 whereas the correct domain is x>0

OpenStudy (er.mohd.amir):

log 0 is not define so x-2 is greater then zero not equal to.

OpenStudy (anonymous):

yes, i corrected that.

OpenStudy (misty1212):

hmmm

OpenStudy (misty1212):

this is not correct

OpenStudy (misty1212):

\[\ln(x-2)\geq 0\] is the first step

OpenStudy (misty1212):

then put in equivalent exponential form as \[x-2\geq e^0\]

OpenStudy (misty1212):

since \(e^0=1\) we now have \[x-2\geq 1 \] solve that in one step

OpenStudy (chillhill):

thank you

OpenStudy (anonymous):

I apologize, didn't see the √ :)

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