Question

Horney little sloot looking for love in all the wrong places!

0=sqrt(x^2+ln(x))

solve for x

then help me with this injective proof

sqrt(x^2+ln(x))=sqrt(y^2+ln(y))

Answer 1

answer the question and ill send you something special ;)

Answer 2

Lol you're a creepy lil lady huh :p
Hmm sorry my brain is too fried today to figure this one out :c

Answer 3

i am what i am.

so here's what i have for this.. really simple actually

For every integers x and y, f(x)=f(y)---> x=y

so,

sqrt(x^2+ln(x))=sqrt(y^2+ln(y))

*end proof*

The problem with the first part is that when trying to solve for x with wolfram i get the attached pic. Problem is is that I am not familiar with complex-valued logs

Answer 4

i found the proof here

http://courses.engr.illinois.edu/cs173/sp2009/lectures/lect_15_supp.pdf

Answer 5

Oh that weird Product-Log thing? I've never really learned about that either.. hmm

Answer 6

i guess ill read up on it then

Answer 7

ya, soz :c

Answer 8

Are you allowed to use a graphing calculator to solve 0=sqrt(x^2+ln(x)) ?

Answer 9

yeah i don't know what to enter though

Answer 10

0=sqrt(x^2+ln(x))
implies x^2+ln(x) = 0

You can plot x^2+ln(x) and find where it crosses the x-axis and that will be your solution OR you can plot y = x^2 and y = -ln(x) and see where they intersect.

Answer 11

err nvm about the graphing party. anyway to do this numerically?

Answer 12

Like Newton-Raphson method?

Answer 13

ive never used it

Answer 14

does the method use complex valued logs?

Answer 15

Which chapter of the lesson does this question appear under. That may give an idea what method they are expecting you to use to solve this.

Answer 16

No, Newton-Raphson method uses calculus. You start with a guess and then you successively approximate a better and better root until you reach the required level of accuracy.

Answer 17

it's somewhere in the transcendental function section..

Answer 18

But they would have discussed various methods for solving transcendental equations. If you list them we can choose the appropriate one to apply here.

Answer 19

You can use a graphing method or use a numerical method. If they use numerical method we need to know what are allowed methods.

But if you want to solve this by trial and error and a calculator I can tell you how.

Answer 20

of ive done the ive just read about as newtons method

Answer 21

I think Newton's Method would work nicely here. It will give you an approximate solution

Answer 22

sweet

Answer 23

I suppose some books call it Newton's Method and some others Newton-Raphson method. Have you covered calculus? Do you know how to find derivatives? Or do you want the trial and error method?

Answer 24

Or do you want the trial and error method without calculus.

Answer 25

yeah i've done it before in calc I but its a little new using natural logs. use calc please

Answer 26

Guess two values for x where x^2+ln(x) changes sign (positive to negative or negative to positive).

We are trying to solve x^2+ln(x) = 0 or x^2 = -ln(x)
The LHS is always positive. Therefore, ln(x) has to be negative for the above equation to work. When is ln(x) negative? ln(x) is zero when x = 1. When x > 1, ln(x) is positive. Therefore, x has to be less than 1 for ln(x) to be negative. And we know ln(x) is not defined for x <=0. Therefore, out x must be 0 < x < 1.

Pick an x value in the interval (0,1). Evaluate f(x).
Pick another number so that you get a change in sign of f(x).
Let me know when you get thiose two x values.

Answer 27

Using your calculator, find f(0.5), f(0.25), f(0.75).

Answer 28

where f(x) = x^2 + ln(x)

Answer 29

.25=-1.324
.5=-.4431
.75=.27482

maybe .6?

Answer 30

Therefore, the root is between x = 0.5 and 0.75
Try the middle value x = (0.5 + 0.75) / 2 = 0.625
(BTW, this trial and error method is called a binary search)

Answer 31

ok i get the point so calc would be maybe .6-(x(x^2+ln(x)))

Answer 32

then i just plug in what i got to the x subscript ns

Answer 33

forever

Answer 34

Binary search using the middle point often leads to the solution faster.
Evaluate f(0.625)
Does it change sign between x = 0.5 and 0.625 or between 0.625 and 0.75?
Then you find the midpoint of that interval and keep going until the desired level of accuracy is reached.

Answer 35

alright, thanks

Answer 36

let us say the desired level of accuracy is two decimal places. Then you keep refining x until two successive solution has the same first two decimals and that is when you stop the iteration.

Answer 37

you are welcome.