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logarithmic functions

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DFranklin

rbnphlp: You seem to be arguing as follows:

I want to solve f(x) = 0. I will multiply both sides by g(x) to get g(x)f(x) = 0. Therefore, any solution to g(x) = 0 is a solution to f(x) = 0.

Put like that, it's obviously invalid reasoning...

Youre right as always .. :eviltongue:

Reply 21

peaches6994

DFranklin

Look guys, I know you want the equation to have a certain form, but I'm really struggling to see how you could *possibly* decide that when someone posts x^2+2x+2ln(x)=0 they really mean

\frac{x^2+2x+1}{2 \ln x} = 0

no these are two different questions, right?

Reply 22

Small123

In terms of specifying 2lnx=0 as a solution it is slightly relevant but whatever, the problems seems to have been resolved.

Reply 23

Ratiasu

rbnphlp

http://www.wolframalpha.com/input/?i=x^2%2B2x%2B2lnx%3D0

that's all well and good but i was hoping someone would tell me how to actually solve it instead of picking the easy way out

Reply 24

qgujxj39

OP, could you actually write the question in a form which isn't ambiguous? I mean, we're 27 posts into the thread and I can't even tell what the exact question is.

Reply 25

rbnphlp

Ratiasu

that's all well and good but i was hoping someone would tell me how to actually solve it instead of picking the easy way out

you should know newton raphson or bisection method something like that to solve it..

Reply 26

Small123

Oh and x=1 isn't a solution :p:

Reply 27

rbnphlp

Small123

Oh and x=1 isn't a solution :p:

yes it is draw a lnx graph (im sure im not wrong this time)

Reply 28

Ratiasu

rbnphlp

you should know newton raphson or bisection method something like that to solve it..

i actually have no idea what you're talking about

oh and i dont know how to write it out 'properly' i've got no programs to do that, can anyone suggest one

Reply 29

Small123

rbnphlp

yes it is draw a lnx graph (im sure im not wrong this time)

substitute x=1 into

\dfrac{x^2+2x+1}{2lnx}

. The answer will not be 0.

Reply 30

rbnphlp

Small123

substitute x=1 into

\dfrac{x^2+2x+1}{2lnx}

. The answer will not be 0.

I was reffering to when lnx=0 , not the question the op asked

Reply 31

peaches6994

small123 just answered you innit? equate the numerator to zero.

Reply 32

Small123

rbnphlp

I was reffering to when lnx=0 , not the question the op asked

Ah, ok.

Reply 33

around

Ratiasu

i actually have no idea what you're talking about

oh and i dont know how to write it out 'properly' i've got no programs to do that, can anyone suggest one

I believe that if you look just above the thread, near where it says 'Useful Resources', there's a link on 'How to use LaTeX'.

either that or just use some damn brackets.

Reply 34

Ratiasu

alright i'll use brackets then

it's (x^2)+(2*x)+2*ln(x)=0 is that any clearer?

Reply 35

qgujxj39

That's very clear, thanks.

We cannot solve this equation to give an exact answer (like 4, 23/2, e^(23 + pi)). Therefore, you're going to have to find an approximate solution using numerical methods. Do you know how to use any of the following?:
- interval bisection
- linear interpolation
- the Newton-Raphson method