1. Tangent line approximation

My problem is $\displaystyle ln(55)$. My answer was $\displaystyle 8/5e^4$. My work is as follows f(x) = ln x
f'(x)= 1/x
55 appx = e^4 which is 54.6
f(e^4)= 4
f'(e^4)= 4/(e^4)
x-a=h= 55 - 54.6 = .4

f(55)= f(55+(55-54.6)) appx = f(e^4) + f'(e^4)*(55 - 54.6)
appx = 4 + 4/(e^4) (55-54.6) = 4 + (8/(e^4)) - 4 = 8/5(e^4)
When you type this in the calculator ln(55) = 4.007 and 8/(5(e^4)) = .029 which is not close at all. Can anyone explain to me what I'm doing wrong, or am missing?

2. Originally Posted by spazzyskylar
My problem is $\displaystyle ln(55)$. My answer was $\displaystyle 8/5e^4$. My work is as follows f(x) = ln x
f'(x)= 1/x
55 appx = e^4 which is 54.6
f(e^4)= 4
f'(e^4)= 4/(e^4)
x-a=h= 55 - 54.6 = .4

f(55)= f(55+(55-54.6)) appx = f(e^4) + f'(e^4)*(55 - 54.6)
appx = 4 + 4/(e^4) (55-54.6) = 4 + (8/(e^4)) - 4 = 8/5(e^4)
When you type this in the calculator ln(55) = 4.007 and 8/(5(e^4)) = .029 which is not close at all. Can anyone explain to me what I'm doing wrong, or am missing?
$\displaystyle f(x) = \ln{x}$

$\displaystyle f'(x) = \frac{1}{x}$

let $\displaystyle x = e^4$ , $\displaystyle y = 4$

tangent line equation at the point $\displaystyle (e^4,4)$ ...

$\displaystyle y - 4 = \frac{1}{e^4}(x - e^4)$

$\displaystyle y - 4 = \frac{x}{e^4} - 1$

$\displaystyle y = \frac{x}{e^4} + 3$

$\displaystyle f(55) \approx \frac{55}{e^4} + 3 = 4.00736013888...$

note that $\displaystyle \ln(55) = 4.00733318523...$