anonymous
  • anonymous
Anyone knows how to solve this??
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\frac{ (x + 5) }{ 3 } = \frac{ \ln(x) }{ -2 }\]
anonymous
  • 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.. ._.
anonymous
  • anonymous
i forget how to do this sry

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mathmate
  • mathmate
hint: Examine what you just wrote: −2x−10=ln(x3)
mathmate
  • mathmate
actually : \(−2x−10=ln(x^3)\)
mathmate
  • mathmate
hint: graph (by hand) each side of the equation \(−2x−10=ln(x^3)\)
anonymous
  • anonymous
well it does imply that e has to have an exponent of -2x-10 to give x^3 so \[e ^{-2x}*\frac{ 1 }{ e ^{10} } = \ln(x^3)\]
mathmate
  • mathmate
|dw:1440154856949:dw| Do they ever intersect?
anonymous
  • anonymous
\[\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\]Choose a suitable approximation.
anonymous
  • 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.
anonymous
  • anonymous
@Audio I'm not sure what that is, can you please expand on that?
anonymous
  • anonymous
It is, indeed, an expansion. :) http://mathworld.wolfram.com/SeriesExpansion.html
anonymous
  • 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.
mathmate
  • mathmate
@nopen Are you working on calculus or numerical methods?
anonymous
  • anonymous
um the working that you see in my first comment is what I used to approach this problem.
mathmate
  • 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.
anonymous
  • 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
mathmate
  • 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.
anonymous
  • anonymous
I'm getting f(0.01) = -4.02 and for f(0.05) = -6.196
anonymous
  • anonymous
but wait how are you taking random numbers for the domain like that?
mathmate
  • 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.
mathmate
  • mathmate
They are not random numbers, they are obtained from the graph that you said earlier that there is a root.
mathmate
  • mathmate
From graphing, we know the value of x is very small, such that -2(x+5) is about -10.
anonymous
  • anonymous
oh. Oops okay then f(0.01) = 3.79 and f(0.05) = -1.11
mathmate
  • mathmate
Anyway, have you recalculated f(0.01) and f(0.05), which should have alternate signs.
mathmate
  • mathmate
Good!
anonymous
  • anonymous
yup
mathmate
  • 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.
mathmate
  • 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)=...
mathmate
  • mathmate
are you still with me?
anonymous
  • anonymous
Yes, but I see that's gonna be a long process by hand :S
mathmate
  • mathmate
Good, now you will appreciate 2. by Newton's method.
mathmate
  • 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
mathmate
  • 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.
anonymous
  • anonymous
Oooh I see what you're doing now
mathmate
  • mathmate
Do you want me to get you started, or that's good enough?
anonymous
  • 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
anonymous
  • anonymous
thanks! @mathmate
mathmate
  • mathmate
If you are curious to use the method, you can check your answer : x=0.03485461440500114
mathmate
  • 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.
anonymous
  • anonymous
lol that's not a few xD
mathmate
  • mathmate
you're welcome! :) by the way, it took 4 iterations, starting from 0.03, the average of 0.01 and 0.05.
anonymous
  • anonymous
oh oops I made a mistake with the starting value
mathmate
  • 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.
anonymous
  • anonymous
ah now I got it right thnx!
mathmate
  • mathmate
Good, have fun with math! :)

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