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Mathematics 7 Online
OpenStudy (amenah8):

find the equation of the tangent line y = ln(x) to the graph at x=e

OpenStudy (anonymous):

start by finding the slope of the tangent line by finding y' at x = e

OpenStudy (amenah8):

the derivative is 1/x

OpenStudy (anonymous):

right, and what's its value when x = e?

OpenStudy (amenah8):

1/e

OpenStudy (amenah8):

1/e

OpenStudy (anonymous):

yeah from here you can use the point slope formula.\[y-y_1=m(x-x_1)\] We found m = 1/3 and were given \(x_1\)= e. Plug in e to the original equation to get \(y_1\)

OpenStudy (amenah8):

one moment while i solve for this please...

OpenStudy (amenah8):

y-lne=1/3(x-e) ?

OpenStudy (anonymous):

sorry I made a typo. 1/3 is supposed to be 1/e. and you can simplify \(\ln e\) to 1

OpenStudy (amenah8):

oh right! so: y-1=1/e(x-e)

OpenStudy (amenah8):

why is the slope 1/e

OpenStudy (anonymous):

because the derivative is the slope of the tangent line. Taking the derivative then plugging in e gave the slope

OpenStudy (amenah8):

ah! i get it! thank you so much

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