As you can check, the point $(1,\ln(2))$ lies on the curve
$xe^y+7y−2x=7\ln(2).$
Find an equation for the straight line tangent to this curve at the given point.
Your answer should be an equation involving x and/or y.
Use ln() for the natural logarithm.
Tried to fin a function y=… but not sure how to do it.
Best Answer
You have a point on the curve: $(x_0, y_0) = (1, \ln 2)$. You need to determine the "slope" $m$ of the tangent line at $x = 1$. To find that slope, you need to find derivative by implicit differentiation: $$\dfrac{2 - e^y}{xe^y + 7}$$ then evaluate at $x = 1, y = \ln 2$, and set $f'(1, \ln 2) = m = 0$.
Then you'll have both a point on the line, and slope, and using point-slope form for an equation, you're good to go:
$$y - y_0 = m(x - x_0)$$
$$y - \ln 2 = 0(x - 1) = 0 \iff y = \ln 2$$ And you've got your line, parallel to the x-axis.