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January 11th, 2017, 01:35 PM   #1
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a problem with ln

So I have this problem that I can't solve.
Can someone help me?

2m + 3 + 2ln|m/(m-1)| = 7 + 2ln(2)

Thank you for helping.

Last edited by skipjack; January 11th, 2017 at 02:09 PM.
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January 11th, 2017, 01:39 PM   #2
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log problem

2m+3+2ln| m/(m-1) | = 7+2ln(2) , m>1

please if possible to help.
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January 11th, 2017, 02:08 PM   #3
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$2m + 3 + 2 \ln\left(\left | \dfrac {m}{m-1}\right | \right)= 7 + 2 \ln(2),~m>1$

as $m>1$ the absolute value signs are unnecessary

$2m -4 = 2\ln(2) - 2 \ln\left(\dfrac{m}{m-1}\right)$

$2(m-2) = 2 \ln\left(\dfrac{2(m-1)}{m}\right)$

$m-2 = \ln\left(\dfrac{2(m-1)}{m}\right)$

In general problems of this sort must be numerically solved.

There is however a simple solution by inspection. I leave it to you to find.

There is another solution that must be found via numeric methods.
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January 11th, 2017, 02:16 PM   #4
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If m has to be an integer, m = 2.
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January 11th, 2017, 02:42 PM   #5
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I got that m=2, but I want to use a method to find it instead of saying it is easy to infer.
Can you give me a clue?

Last edited by skipjack; January 12th, 2017 at 05:12 AM.
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January 11th, 2017, 02:47 PM   #6
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Quote:
Originally Posted by nuaheuc View Post
I got that m=2, but I want to use a method to find it instead of saying it is easy to infer.
Can you give me a clue?
As I said, you have to use numeric methods.

The first thing to do is plot the difference of the two sides.

You can see from that graph that solutions are at approximately 1.35 and 2.

You can immediately plug 2 in and see that it is in fact a solution.

You have to use something like Newton's method with a starting point of 1.35 or so to obtain the second solution.

This is the best you can do.

Last edited by skipjack; January 12th, 2017 at 05:13 AM.
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January 11th, 2017, 04:32 PM   #7
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Quote:
Originally Posted by nuaheuc View Post
I got that m=2, but I want to use a method to find it instead of saying it is easy to infer.
Can you give me a clue?
Sort of.

$m > 1\ and\ 2m + 3 + 2\ln \left | \dfrac{m}{m - 1} \right | = 7 + 2 \ln(2) \implies$

$m + \ln \left ( \dfrac{m}{2(m - 1)} \right) = 2 \implies \ln \left ( \dfrac{me^m}{2(m - 1)} \right) - 2 = 0.$

You can now graph the function $f(m) = \left ( \dfrac{me^m}{2(m - 1)} \right) - 2.$

This graph will indicate APPROXIMATELY where the zeros of the function are.

One of those zeros will appear to equal 2. It is then easy to verify that m = 2 is an exact answer. There may be other answers. The graph will then indicate where you can use numerical methods such as Newton-Raphson to get approximations for those other zeros to any desired degree of accuracy.

Last edited by skipjack; January 12th, 2017 at 05:14 AM.
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January 11th, 2017, 07:49 PM   #8
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Hello nuaheuc. Please do not post the same question in two different threads. This clutters the forum and creates redundancy.

In regards to your problem, what is m if 2m + 3 = 7?
What is m if m/(m - 1) = 2? Do the two solutions concur?
If so, you have found a solution.
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