How do you find the equation of the tangent line to the graph #y=log_3x# through point (27,3)?

1 Answer
Shwetank Mauria
May 22, 2017

#x-27ln3-27+81ln3=0#

Explanation:

As slope of tangent to a curve at a given point is give by the value of first derivative at that point, let us first find the differential of #y=log_3x#.

As #y=log_3x=lnx/ln3#

#(dy)/(dx)=1/(xln3)# and at #(27,3)# slope is #1/(27ln3)#

Hence equation of tangent is

#y-3=1/(27ln3)(x-27)#

or #27ln3y-81ln3-x+27=0#

or #x-27ln3-27+81ln3=0#

graph{(x-27ln3y-27+81ln3)(y-lnx/ln3)((x-27)^2+(y-3)^2-0.05)=0 [-10, 30, -10, 10]}

Related questions
See all questions in Differentiating Exponential Functions with Other Bases
Impact of this question
3703 views around the world
You can reuse this answer
Creative Commons License