Consider the curve y=ln(3x-1).let p be the point on the curve where x=2.
a. a write down the gradient of the curve at p.
b. the normal to the curve at p cuts the x axis at r. find the coordinate of r. (is the gradient of p 3/(3x-1)?

A) First off:

For y = Inx    dy/dx  =  1/x

For y = In(3x-1)    dy/dx  =  1/(3x-1)  * 3  = 3/(3x-1).

So yes the gradient at P is =  3/(3x-1)

But also remember at P,  x = 2.     Therefore gradient =    3 / (3*2 -1)   = 3 /(6-1)

Gradient, m at P is actually =  3 /5 = 0.6

b) The normal to curve:

m1 * m2 = -1.    Condition for normal (perpendicular)

m2  =  -1  / (3/5)  = -5/3

Recall  at point P,  x =2,  y = In(3x-1)   = In(3*2-1) = ln5

Hence Point P   =              ( 2, In5)

Equation of gradient and one point:

y - y1 =  m(x - x1).              Here  x1 = 2,      y1 = In5

y - ln5  =  (-5/3)(x  - 2)

3y - 3ln5  = -5x + 10

3y =  -5x + 10 + 3ln5    ,  3y =  -5x + 10 + ln5^3

3y =  -5x + 10 + ln125

Hence normal to the curve is      3y =  -5x + (10 + ln125)

When normal cuts the x axis, at that point y = 0. 

3y =  -5x + (10 + ln125)

3*0 =  -5x + (10 + ln125)

0 =  -5x + (10 + ln125)

5x =  (10 + ln125)    Divide both sides by 10

x  =  (10 + ln125) / 5  = 2.9657.

Coordinates of R  =  ( 2.97, 0).

It will take him 5 days

The answer is $9.90 with rounding to the nearest tenth