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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.

Lain Symon
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1 Answers1

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$\ln{y} = 3x+1 \implies y = e^{3x+1}\\y' = 3e^{3x+1} \implies y'(0) = 3e\tag*{}$

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 ...

$y - e = 3e(x - 0)\tag*{}$

Deepak Suwalka
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