# Thread: linear approximation... of ln(1.01)

1. ## linear approximation... of ln(1.01)

i pretty much understand linear approximations but i cant seem to solve this problem. if anyone can show me some steps to get me started i would really love that
Use linear approximation to approximate the number ln(1.01)

we know that
ln1 = 0

2. Originally Posted by delcidm7
i pretty much understand linear approximations but i cant seem to solve this problem. if anyone can show me some steps to get me started i would really love that
Use linear approximation to approximate the number ln(1.01)

we know that
ln1 = 0
yes, we do know that

let $f(x) = \ln x$

use the linear approximation formula: $f(x) \approx f(a) + f'(a)(x - a)$

here, x = 1.01, a = 1

3. ok so i got
1st: $f(1.01) \approx f(1) + f'(1)(1.01 - 1)$
then: $f(1.01) \approx 0 + 1(.01)$
finally: $f(1.01) \approx (.01)$
thanks so much!!

4. Originally Posted by delcidm7
ok so i got

$
f(1.01) \approx f(1) + f'(1)(1.01 - 1)
$

$
f(1.01) \approx 0 + 1(.01)$

$
f(1.01) \approx (.01)$

thanks so much!!
correct