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November 26th, 2018, 03:50 AM  #1 
Newbie Joined: Nov 2018 From: RU Posts: 5 Thanks: 0  Help, beg you. Equation with parameter
Sqrt (56*x)*ln(4*sqr(x)sqr(a))=sqrt(56*x)*ln(2*x+a) Find all possible a when an equation has only one possible solution. Отправлено с моего SMA750FN через Tapatalk 
November 26th, 2018, 04:47 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,310 Thanks: 1980 
A solution other than x = 5/6 must satisfy ln(4√x  √a) = ln(2*x+a). Can you make any progress from there? That would be "must satisfy ln(4x²  a²) = ln(2*x+a)" if "sqr" meant"²". Last edited by skipjack; November 26th, 2018 at 11:27 PM. 
November 26th, 2018, 04:54 AM  #3 
Newbie Joined: Nov 2018 From: RU Posts: 5 Thanks: 0 
I can't, but I need to solve it Отправлено с моего SMA750FN через Tapatalk 
November 26th, 2018, 06:00 AM  #4  
Senior Member Joined: Sep 2016 From: USA Posts: 564 Thanks: 326 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
 
November 26th, 2018, 08:03 AM  #5  
Senior Member Joined: May 2016 From: USA Posts: 1,307 Thanks: 549  Quote:
does sqr(x) mean x^2 = $x^2$? And is the case of x = 5/6 relevant, and if not, why not?  
November 26th, 2018, 08:05 AM  #6 
Newbie Joined: Nov 2018 From: RU Posts: 5 Thanks: 0 
Y, it means x^2 Отправлено с моего SMA750FN через Tapatalk 
November 26th, 2018, 11:06 AM  #7 
Newbie Joined: Nov 2018 From: RU Posts: 5 Thanks: 0 
5/6 is relevant, but we need a. I am confused Отправлено с моего SMA750FN через Tapatalk 
November 26th, 2018, 11:41 AM  #8 
Newbie Joined: Nov 2018 From: RU Posts: 5 Thanks: 0 
4x^2a^2  2x  a = 0 So we need to find a values when D=0? And if a=1, then it's quadrant equation? Отправлено с моего SMA750FN через Tapatalk 
November 27th, 2018, 03:51 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,310 Thanks: 1980 
If 0 = 4x²  a²  2x  a = (2x + a)(2x  a  1), 2x + a = 0 (consider when this is good) or x = (a + 1)/2 (consider when this is good). 

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