Question

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

Answer 1

`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]}