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Mathematics 11 Online
OpenStudy (anonymous):

ln sqrt(x+2) = 1

OpenStudy (anonymous):

Square both side, what do you get?

OpenStudy (anonymous):

ln x+2 = 1 ?

hartnn (hartnn):

ln is log to base 'e'

OpenStudy (anonymous):

Yes, my mistake! :(

OpenStudy (anonymous):

Right, so I need to solve for x. I believe its x = e\[^{2}\] - 2

OpenStudy (anonymous):

x = e^2-2

OpenStudy (anonymous):

Yes :)

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