January 11th, 2017, 12:35 PM  #1 
Newbie Joined: Jan 2017 From: israel Posts: 3 Thanks: 0  a problem with ln
So I have this problem that I can't solve. Can someone help me? 2m + 3 + 2lnm/(m1) = 7 + 2ln(2) Thank you for helping. Last edited by skipjack; January 11th, 2017 at 01:09 PM. 
January 11th, 2017, 12:39 PM  #2 
Newbie Joined: Jan 2017 From: israel Posts: 3 Thanks: 0  log problem
2m+3+2ln m/(m1)  = 7+2ln(2) , m>1 please if possible to help. 
January 11th, 2017, 01:08 PM  #3 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,493 Thanks: 752 
$2m + 3 + 2 \ln\left(\left  \dfrac {m}{m1}\right  \right)= 7 + 2 \ln(2),~m>1$ as $m>1$ the absolute value signs are unnecessary $2m 4 = 2\ln(2)  2 \ln\left(\dfrac{m}{m1}\right)$ $2(m2) = 2 \ln\left(\dfrac{2(m1)}{m}\right)$ $m2 = \ln\left(\dfrac{2(m1)}{m}\right)$ 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. 
January 11th, 2017, 01:16 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,048 Thanks: 1395 
If m has to be an integer, m = 2.

January 11th, 2017, 01:42 PM  #5 
Newbie Joined: Jan 2017 From: israel Posts: 3 Thanks: 0 
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 04:12 AM. 
January 11th, 2017, 01:47 PM  #6  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,493 Thanks: 752  Quote:
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 04:13 AM.  
January 11th, 2017, 03:32 PM  #7  
Senior Member Joined: May 2016 From: USA Posts: 802 Thanks: 318  Quote:
$m > 1\ and\ 2m + 3 + 2\ln \left  \dfrac{m}{m  1} \right  = 7 + 2 \ln(2) \implies$ $m + \ln \left ( \dfrac{m}{2(m  1)} \right) = 2 \implies \ln \left ( \dfrac{me^m}{2(m  1)} \right)  2 = 0.$ 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 NewtonRaphson to get approximations for those other zeros to any desired degree of accuracy. Last edited by skipjack; January 12th, 2017 at 04:14 AM.  
January 11th, 2017, 06:49 PM  #8 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,593 Thanks: 937 Math Focus: Elementary mathematics and beyond 
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. 