Find the equation of the tangent to the curve y=ln(3x-2) + 4 at the point where x=1?

Thanks!

Nov 27, 2017

$y = 3 x + 1$

Explanation:

•color(white)(x)m_(color(red)"tangent")=dy/dx" at x = a"

$\text{differentiate using the "color(blue)"chain rule}$

$\text{given "y=f(g(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \text{chain rule}$

$y = \ln \left(3 x - 2\right) + 4$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3 x - 2} \times \frac{d}{\mathrm{dx}} \left(3 x - 2\right) = \frac{3}{3 x - 2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(x = 1\right) = \frac{3}{1} = 3$

$y = \ln \left(3 - 2\right) + 4 = 0 + 4 = 4 \Rightarrow \left(1 , 4\right)$

$\Rightarrow y - 4 = 3 \left(x - 1\right) \leftarrow \text{point-slope form}$

$\Rightarrow y = 3 x + 1 \leftarrow \text{slope-intercept form}$
graph{(y-ln(3x-2)-4)(y-3x-1)=0 [-20, 20, -10, 10]}