Find the equation of the tangent to the curve y = ln(3x − 2) at the point where x = 1. I differentiated this and got 3/3x-2, whats my next step?
to find the equation of a line tangent to a curve defined by $\displaystyle f(x)$ at $\displaystyle x = a$ ...
1. you need the point of tangency on the curve, $\displaystyle (a, f(a))$
2. you need the value of the slope of the curve at the point of tangency , $\displaystyle m = f'(a)$
3. now that you have a point $\displaystyle (a,f(a))$ and a slope, $\displaystyle m = f'(a)$ , write the equation in point-slope form ...
$\displaystyle y - f(a) = f'(a)(x - a)$