Find the equation of tangent to the curve y = ln x at the point where x = a. Hence, find the equation of tangent to the curve which passes through the origin. Given that line y = mx intersects the curve the curve y = ln x at two distinct pts, write down an inequality for m.

Question

amistre64 · Answer

what do we get for the derivative?

amistre64 · Answer

the derivative defines the slope at any given value of x; therfore the tangent line at any given points is defined as:

y = f'(a) (x-a) + f(a)

amistre64 · Answer

then your next task seems to define the range of value for m such that lnx = mx has only 2 solutions

anonymous · Answer

Yes and then???

amistre64 · Answer

and then you come up with the results ...

anonymous · Answer

I still dont get the first part.

amistre64 · Answer

then ask questions about it; what dont you understand?

anonymous · Answer

I differentiate ln x i get 1/x

amistre64 · Answer

good,  and 1/x at any given x=a is just 1/a

anonymous · Answer

Correct

amistre64 · Answer

y = 1/a (x-a) + ln(a)  is the slope of the tangent line

amistre64 · Answer

lol, equation of the tangent line

anonymous · Answer

*x-a + ln a * i see

amistre64 · Answer

when does this equation have an x and y intercept at the origin?

anonymous · Answer

Lol i only know that method

amistre64 · Answer

in other words, for what value of a is (0,0) a good solution to it?

amistre64 · Answer

y = 1/a (x-a) + ln(a)

0 = 1/a (0-a) + ln(a)

0 = -1 + ln(a)

1 = ln(a) , when a=e

amistre64 · Answer

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