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November 26th, 2018, 03:50 AM   #1
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Help, beg you. Equation with parameter

Sqrt (5-6*x)*ln(4*sqr(x)-sqr(a))=sqrt(5-6*x)*ln(2*x+a)

Find all possible a when an equation has only one possible solution.

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November 26th, 2018, 04:47 AM   #2
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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.
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November 26th, 2018, 04:54 AM   #3
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I can't, but I need to solve it

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November 26th, 2018, 06:00 AM   #4
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Originally Posted by Alvalance789 View Post
I can't, but I need to solve it

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November 26th, 2018, 08:03 AM   #5
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Quote:
Originally Posted by Alvalance789 View Post
Sqrt (5-6*x)*ln(4*sqr(x)-sqr(a))=sqrt(5-6*x)*ln(2*x+a)

Find all possible a when an equation has only one possible solution.

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As you were asked at free math help,

does sqr(x) mean x^2 = $x^2$?

And is the case of x = 5/6 relevant, and if not, why not?
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November 26th, 2018, 08:05 AM   #6
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Y, it means x^2

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November 26th, 2018, 11:06 AM   #7
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5/6 is relevant, but we need a. I am confused

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November 26th, 2018, 11:41 AM   #8
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4x^2-a^2 - 2x - a = 0

So we need to find a values when D=0?

And if a=1, then it's quadrant equation?

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November 27th, 2018, 03:51 AM   #9
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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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