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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ (x + 5) }{ 3 } = \frac{ \ln(x) }{ 2 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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
 one year ago
Best ResponseYou've already chosen the best response.0i forget how to do this sry

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4hint: Examine what you just wrote: −2x−10=ln(x3)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4actually : \(−2x−10=ln(x^3)\)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4hint: graph (by hand) each side of the equation \(−2x−10=ln(x^3)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well it does imply that e has to have an exponent of 2x10 to give x^3 so \[e ^{2x}*\frac{ 1 }{ e ^{10} } = \ln(x^3)\]

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4dw:1440154856949:dw Do they ever intersect?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\ln(1+x) = x  \frac{x^2}{2} + \frac{x^3}{3}  \frac{x^4}{4} + \cdots\]Choose a suitable approximation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.0@Audio I'm not sure what that is, can you please expand on that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It is, indeed, an expansion. :) http://mathworld.wolfram.com/SeriesExpansion.html

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.4@nopen Are you working on calculus or numerical methods?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0um the working that you see in my first comment is what I used to approach this problem.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4This 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
 one year ago
Best ResponseYou've already chosen the best response.0ah, 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
 one year ago
Best ResponseYou've already chosen the best response.4The 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
 one year ago
Best ResponseYou've already chosen the best response.0I'm getting f(0.01) = 4.02 and for f(0.05) = 6.196

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but wait how are you taking random numbers for the domain like that?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4Sorry, 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
 one year ago
Best ResponseYou've already chosen the best response.4They are not random numbers, they are obtained from the graph that you said earlier that there is a root.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4From graphing, we know the value of x is very small, such that 2(x+5) is about 10.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh. Oops okay then f(0.01) = 3.79 and f(0.05) = 1.11

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4Anyway, have you recalculated f(0.01) and f(0.05), which should have alternate signs.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4Now, 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
 one year ago
Best ResponseYou've already chosen the best response.4For 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
 one year ago
Best ResponseYou've already chosen the best response.4are you still with me?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, but I see that's gonna be a long process by hand :S

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4Good, now you will appreciate 2. by Newton's method.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.42. 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=x0f(x0)/f'(x0) where f'(x)=d f(x)/d x = 2  3/x in this case

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4and 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
 one year ago
Best ResponseYou've already chosen the best response.0Oooh I see what you're doing now

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4Do you want me to get you started, or that's good enough?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4If you are curious to use the method, you can check your answer : x=0.03485461440500114

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4If 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
 one year ago
Best ResponseYou've already chosen the best response.0lol that's not a few xD

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4you're welcome! :) by the way, it took 4 iterations, starting from 0.03, the average of 0.01 and 0.05.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh oops I made a mistake with the starting value

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4Exactly, 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
 one year ago
Best ResponseYou've already chosen the best response.0ah now I got it right thnx!

mathmate
 one year ago
Best ResponseYou've already chosen the best response.4Good, have fun with math! :)
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