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Use differentials (or equivalently, a linear approximation) to approximate ln(0.91) as follows: Let ƒ(x)=ln(x) and find the equation of the tangent line to ƒ(x) at a "nice" point near 0.91. Then use this to approximate ln(0.91). Approximation= ?
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i got the answer which is -0.09...anyway you can show how you got it? I did it, but i'm pretty sure there's different ways to do it...
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