Question

f(x)=ln3x has a tangent line at point x=1/3 , find point (x;29)

Answer 1

the instructions are a bit vague

Answer 2

I don't really know the correct translation . But I'll try to make something else out of it .

Answer 3

f(x)=ln3x has a tangent line at point x=1/3 , find point on the tangent line , where y=29 .

Answer 4

I thought about first finding y , which of 0 , then derivative 1/x , then slope , which is 3 and then using formula k=y2-y1/x2-x1 .

Answer 5

ln(3*1/3) = ln(1) = 0

thats a good thought ... lets see if we can refine it

Answer 6

ln(3x) derives to: 3/3x = 1/x

the slope of the line at x=1/3 is then 3

y = mx -mPx + Py is the equation of a line:

siince our m=3; and our (Px,Py) is (1/3, 0)

29 = 3x -3(1/3) + 0

29 = 3x -1

3x = 30, x=10

Answer 7

the formula you proposed for "k" is just the slope formula; we needed the slope intercept form of a line to describe the tangent line itself

Answer 8

Thanks!

Answer 9

youre welcome