I am trying to solve for the tangent line to a curve at a given point. The exact problem is:
$\ln(y) = 3x+1$ at the point $(0,e)$
The first thing I did was solve for $y$ by raising both sides of the equation to the $e$
$$e^{\ln(y)} = e^{3x+1}\\y = e^{3x+1}$$
Second thing I did was find the derivative of
$y = e^{3x+1}.$ I got $y' = 3e^{3x+1}$
Assuming the above steps are correct I am not quite sure where to go from here?
Thanks in advanced for any help you can give me.
$\ln{y} = 3x+1 \implies y = e^{3x+1}\\y' = 3e^{3x+1} \implies y'(0) = 3e\tag*{}$
at the point, $(0,e)$, tangent line can be determined using the point-slope form ...
$y - e = 3e(x - 0)\tag*{}$
method 2
$\dfrac{\mathrm d}{\mathrm dx}\left(\ln{y} = 3x+1 \right)\tag*{}$
implicit derivative ...
$\dfrac{y'}{y} = 3\\y' = 3y\tag*{}$
at the point, $(0,e)$, $y' = 3e$
tangent line in point-slope form ...
$y - e = 3e(x - 0)\tag*{}$