Linearization of ln(7x) at a=1/7

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SUMMARY

The linearization of the function f(x) = ln(7x) at the point a = 1/7 is derived using the formula L(x) = f(a) + f'(a)(x - a). The calculated value f(1/7) equals 0, while the derivative f'(1/7) is confirmed to be 9.12. Therefore, the correct linearization is L(x) = 0 + 9.12(x - 1/7), which simplifies to L(x) = 9.12x - 1.306. The discussion emphasizes the importance of correctly applying the properties of logarithms, specifically ln(ax) = ln(a) + ln(x), to find the derivative accurately.

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Homework Statement


find the linearization of L(x) at a.
f(x)=ln7x, a=1/7


Homework Equations


f(a)+f'(a)(x-a)


The Attempt at a Solution



i got f(1/7)=0 and f'(1/7)=9.12
then shouldn't it be 0+.9.12(x-0)=9.12x?
 
Last edited by a moderator:
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Not quite - Check your derivative again - if it helps, remember, ln(ax) = ln(a) + ln(x).
 

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