# Need help with a nooby question

1. Sep 18, 2006

### semc

alright there is this question which uses NR method to find intersect but i kept getting the wrong ans can anybody help me check which part i went wrong?

f(x)=2x + ln4x , f'(x)=2+ 1/x

btw,differentiate ln4x and lnx get the same result but the arbitary constant is different right?so do we consider the constant when finding the intersect?

Last edited: Sep 18, 2006
2. Sep 18, 2006

If you integrate 1/x, you get lnx + C. ln4x = ln4 + lnx, so C = ln4 in this case.

3. Sep 18, 2006

### semc

hmm...i see now so the constant changes although theres 1/x in both cases.

regarding the question, if i were to find the 2nd approx. , i am suppose to use 0.1 as 1st approx. however, 0.1-f(0.1)/f'(0.1) is not the ans

4. Sep 19, 2006

### HallsofIvy

Staff Emeritus
Why not? It won't be the exact solution to the equation, of course, but it certainly is a "second approximation" with first approximation 0.1. I get 0.16 as the second approximation.

By the way, the reason ln(4x) and ln(x) have the same derivative (1/x) is that ln(4x)= ln(x)+ ln(4).

5. Sep 19, 2006

### semc

hmm....i also get 0.16 guess the ans is wrong Thanks