Anyone knows how to solve this??

- anonymous

Anyone knows how to solve this??

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- anonymous

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

- 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.. ._.

- anonymous

i forget how to do this sry

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## More answers

- mathmate

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

- mathmate

actually : \(−2x−10=ln(x^3)\)

- mathmate

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

- 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

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

- anonymous

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

- 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.

##### 1 Attachment

- anonymous

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

- anonymous

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

- 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

@nopen Are you working on calculus or numerical methods?

- anonymous

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

- 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

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

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

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

- anonymous

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

- 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

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

- mathmate

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

- anonymous

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

- mathmate

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

- mathmate

Good!

- anonymous

yup

- 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

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

are you still with me?

- anonymous

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

- mathmate

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

- 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

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

Oooh I see what you're doing now

- mathmate

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

- 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

thanks! @mathmate

- mathmate

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

- 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

lol that's not a few xD

- mathmate

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

- anonymous

oh oops I made a mistake with the starting value

- 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

ah now I got it right
thnx!

- mathmate

Good, have fun with math! :)

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