## anonymous one year ago Anyone knows how to solve this??

1. anonymous

$\frac{ (x + 5) }{ 3 } = \frac{ \ln(x) }{ -2 }$

2. anonymous

I'm really confused how to go on about this. I tried doing it like this: $-2(x + 5) = 3\ln(x)$ $-2x -10 = \ln(x^3)$ $e ^{-2x -10} = x^3$ and after I'm just stuck.. ._.

3. anonymous

i forget how to do this sry

4. mathmate

hint: Examine what you just wrote: −2x−10=ln(x3)

5. mathmate

actually : $$−2x−10=ln(x^3)$$

6. mathmate

hint: graph (by hand) each side of the equation $$−2x−10=ln(x^3)$$

7. anonymous

well it does imply that e has to have an exponent of -2x-10 to give x^3 so $e ^{-2x}*\frac{ 1 }{ e ^{10} } = \ln(x^3)$

8. mathmate

|dw:1440154856949:dw| Do they ever intersect?

9. anonymous

$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$Choose a suitable approximation.

10. anonymous

@mathmate yes they do intersect which means that they have one 1 root but I want to find the value for x in order to find the coordinate of the intersection.

11. anonymous

@Audio I'm not sure what that is, can you please expand on that?

12. anonymous

It is, indeed, an expansion. :) http://mathworld.wolfram.com/SeriesExpansion.html

13. anonymous

The best bet here is to graph though. You never know how inaccurate an answer can be when you use Taylor Expansions to solve transcendental equations.

14. mathmate

@nopen Are you working on calculus or numerical methods?

15. anonymous

um the working that you see in my first comment is what I used to approach this problem.

16. mathmate

This is basically a numerical methods problem, in which you know there is a root, and need a method to refine the value accurately. This is the approach.

17. anonymous

ah, yes that's exactly what I wanted to do actually. But I don't know really know how, as I'm stuck at the 3rd step

18. mathmate

The third step is the meat of the problem. start with something simple,like f(x)=-2(x+5)-3log(x) and calculate f(0.01) and f(0.05) to confirm that there is a root.

19. anonymous

I'm getting f(0.01) = -4.02 and for f(0.05) = -6.196

20. anonymous

but wait how are you taking random numbers for the domain like that?

21. mathmate

Sorry, I usually use log(x) to mean ln(x), and log10(x) when it's to base 10. We use log10 so rarely that I use log to mean log e.

22. mathmate

They are not random numbers, they are obtained from the graph that you said earlier that there is a root.

23. mathmate

From graphing, we know the value of x is very small, such that -2(x+5) is about -10.

24. anonymous

oh. Oops okay then f(0.01) = 3.79 and f(0.05) = -1.11

25. mathmate

Anyway, have you recalculated f(0.01) and f(0.05), which should have alternate signs.

26. mathmate

Good!

27. anonymous

yup

28. mathmate

Now, we can do one of the two things. 1. by trial and error, or bisection method: Keep calculating f(x) while narrowing down between 0.01 and 0.05 until you get f(x) as close to 0 as you wish. You'd better have a programmable calculator or a computer for that.

29. mathmate

For example, f(0.01)=3.795510557964274 f(0.05)=-1.112803179338027 f(0.035)=-0.01277834752183082 make guesses along the way f(0.03465)=0.01807266003867447 f(0.03475)=...

30. mathmate

are you still with me?

31. anonymous

Yes, but I see that's gonna be a long process by hand :S

32. mathmate

Good, now you will appreciate 2. by Newton's method.

33. mathmate

2. Newton's method says given f(x)=-2(x+5)-3log(x) and an initial approximation x0, under certain conditions of convergence (which we satisfy here), a better approximation x1=x0-f(x0)/f'(x0) where f'(x)=d f(x)/d x = -2 - 3/x in this case

34. mathmate

and we can repeat the process until $$x_n$$ is identical to $$x_{n+1}$$. The best part is that the number of correct decimal places doubles after each iteration because it has second order convergence.

35. anonymous

Oooh I see what you're doing now

36. mathmate

Do you want me to get you started, or that's good enough?

37. anonymous

Oh that's good enough thank you very much, because the original question just asked to mention if there is any roots if so state the number of them, so I was just curious to find the coordinate for it

38. anonymous

thanks! @mathmate

39. mathmate

If you are curious to use the method, you can check your answer : x=0.03485461440500114

40. mathmate

If you are curious to use the method, you can check your answer : x=0.03485461440500114 accurate to the last digit, after just a few iterations.

41. anonymous

lol that's not a few xD

42. mathmate

you're welcome! :) by the way, it took 4 iterations, starting from 0.03, the average of 0.01 and 0.05.

43. anonymous

oh oops I made a mistake with the starting value

44. mathmate

Exactly, the key is to find a good starting value! But any one reasonably close to the final root is ok, depending on the problem, and the conditions of convergence.

45. anonymous

ah now I got it right thnx!

46. mathmate

Good, have fun with math! :)