Find the equation of the tangent line to y=lnx at x=1
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Find the equation of the tangent line to y=lnx at x=1

[From: ] [author: ] [Date: 11-10-19] [Hit: ]
Use the approximation y(x) = x - 1 ≈ ln(x) to evaluate y(1.2) and y(2.5). (Just plug those values into the line).......
I got the equation and its x-1, but I don't know how to answer the second problem. Use it to calculate approximate values for ln(1.2) and ln(2.5). How do I calculate the approximate values? .

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y(x) = ln(x)
y(1) = 0

y'(x) = 1/x
y'(1) = 1

Point-Slope Form:
y = 0 + 1(x - 1)
y = x - 1

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Use the approximation y(x) = x - 1 ≈ ln(x) to evaluate y(1.2) and y(2.5). (Just plug those values into the line).
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