# Math Help - [SOLVED] Finding the equation of curve

1. ## [SOLVED] Finding the equation of curve

The gradient of the tangent to a curve at the point (x,y) where x>3, is inversely proportional to (x-3), and the curve passes through the points (4,0) and (6,ln9). Show that the equation of the curve is y=2ln(x-3).

Thanks

2. Originally Posted by stpmmaths
The gradient of the tangent to a curve at the point (x,y) where x>3, is inversely proportional to (x-3), and the curve passes through the points (4,0) and (6,ln9). Show that the equation of the curve is y=2ln(x-3).

Thanks
If $f'(x) = \dfrac k{x-3}$ then $f(x)= k \cdot \ln(x-3)+C$

$(4, 0)$ satisfies the equation:

$0 = k\ln(4-3)+C~\implies~C=0$

$(6, \ln(9))$ satisfies the equation:

$\ln(9)= k \cdot \ln(6-3)~\implies~k=\dfrac{\ln(9)}{\ln(3)}=2$

3. It's that easy only. Thanks.