Show that the tangent to the curve y=lnx at the point (e,1) also passes through (0,0)?

Answer 1

Here you need calculus-level math knowledge to solve it.

First of all, what's the SLOPE of this function at that point?

If y = ln x, then the slope at any point from calculus theory is dy/dx = 1/x. [It's not a linear function, but tables of derivatives will give that to you, and the proof is there somewhere in your calculus textbook.]

At the point (e,1) the slope is 1/e. But then this is m in the equation of the line y = mx + b, so solve for b since you know the line includes (e,1):

1 = (1/e)e + b
1 = 1 + b
b = 0

So the line is y = (1/e)x and since b = 0 you know the origin (0,0) must be included on the line. Ta-daa!

Answer 2

y=ln(x) is equivalent to: x=e(y).

The slope of x=e(y) is e(y) as well...by definition.

at the point (1,e)...the slope = e

thus...the tangent (slope) is e.

a straight line with a slope of e at point 1,e of course passes through the origin...

(1-0)*e=(e-0)

Answer 3

f(x) = ln x
f `(x) = 1 / x
f `(e) = 1 / e
Equation of tangent thro` (e,1) is:-
y - 1 = (1/e).(x - e)
y = (1/e).x
This line satisfies point (0,0)