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The point P(2,ln(2)) lies on the curve y=ln(x). (a) If Q is the point (x,ln(x), find the slope of the secant line PQ correct to four decimal places for the following values of x:

(1) 1.5 (2) 1.9 (3) 1.99 (4) 1.999 (5) 2.5 (6) 2.1 (7) 2.01 (8) 2.001

(b) Guess the slope of the tangent line to the curve at P. (c) Using the slope from part (b), find the equation of the tangent line to the curve at P.

I have already found the slope of the secant line for each of the values, and guessed the slope of the tangent line to be 0.5. I am stuck on how to write the equation, I am sure I am overthinking it but is't the slope going to be 0.5? How do I find the y-intercept?

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Once you have found the slope of the line tangent to the curve at a given point $(x_1, y_1)$, you can simply use the point-slope form of a line $y-y_1=m(x-x_1)$ for the equation of your tangent line at this point. In your case, $y-ln(2)=\frac{1}{2}(x-2)$.

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If you think that the slope is $\frac 12$, then the equation of the tangent line looks like $$y = \frac12 x + b.$$

If you know the $x$ and $y$ values of a single point on that line (and you do), you can compute $b$, which is the $y$-intercept: $$y(0) = \frac12 \cdot 0 + b= b.$$

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