√(5-6*x)*ln(4*√(x)-√(a))=√(5-6*x)*ln(2*x+a)

[Avalance789]

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.

 

[HOI]

First we have to assume that x= 5/6 is not a solution since that would make both sides 0 no matter what a is. In that case, we can divide both sides by "Sqrt (5-6*x)" to make the equation ln(4*sqr(x)-sqr(a))=ln(2*x+a), Since ln is a one-to-one function we must have 4\sqrt{x}- \sqrt{a}= 2x+ a. Let y= \sqrt{x} so that y^2= x. The equation becomes 4y- \sqrt{a}= 2y^2+ a or 2y^2- 4y+ a- \sqrt{a}= 0. Since y= \sqrt{a}, y must be positive. a must be such that the quadratic equation 2y^2- 4y+ a- \sqrt{a}= 0 has only one positive root. Use the quadratic formula to determine what a must be.

 

[Avalance789]

I am sorry, but I am unable to determine it. But if I won't solve it, I am dead man(

 

[Greg]

What do you know about quadratic equations?

$$ax^2+bx+c=0\implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

The desired quadratic has only one real root so you are looking for $b^2-4ac=0$ (with $a$ as in the equation in this post). Apply that to the quadratic Country Boy gave.

When summarising the given problem I'd pay more attention to the trivial solution mentioned by Country Boy.

 

[HOI]

By the quadratic formula, which greg1313 wrote out for you, the solutions to the quadratic equation [math]2y^2- 4y+ a- \sqrt{a}= 0[/math] are given by

[math]\frac{4\pm\sqrt{16- 8(a-\sqrt{a})}}{4}[/math].

Clearly taking the "+" sign gives a positive root. In order that there be only one positive root we must have [math]4-\sqrt{16- 8(a- \sqrt{a})}< 0[/math]. From that, [math]\sqrt{16- 8(a- \sqrt{a})}> 4[/math] and, squaring both sides, [math]16- 8(a- \sqrt{a})> 16[/math] so [math]8(a- \sqrt{a})< 0[/math], [math]a< \sqrt{a}[/math].

Now, what must \(a\) be?

 

[Avalance789]

So it doesn't have solution? My teacher states on that it has

 

[Avalance789]

Forces me to solve it ultimatively

 

[Avalance789]

Teacher gives me result

(-5/3; -1/2] [2/3; 5/3)

Is it correct?

 

[Avalance789]

Sorry, I have confused you. Should be written like

sqrt (5-6x)*ln(4x^2-a^2)=sqrt(5-6x)*ln(2x+a)So sorry

 

[Avalance789]

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?

 

[MarkFL]

Sorry, I have confused you. Should be written like

sqrt (5-6x)*ln(4x^2-a^2)=sqrt(5-6x)*ln(2x+a)So sorry

Okay, so the equation is:

$$\sqrt{5-6x}\ln\left(4x^2-a^2\right)=\sqrt{5-6x}\ln\left(2x+a\right)$$

Let's arrange as:

$$\sqrt{5-6x}\ln\left(4x^2-a^2\right)-\sqrt{5-6x}\ln\left(2x+a\right)=0$$

Factor:

$$\sqrt{5-6x}\left(\ln\left(4x^2-a^2\right)-\ln\left(2x+a\right)\right)=0$$

Apply log rule for subtraction:

$$\sqrt{5-6x}\ln\left(\frac{4x^2-a^2}{2x+a}\right)=0$$

Factor log argument:

$$\sqrt{5-6x}\ln\left(\frac{(2x+a)(2x-a)}{2x+a}\right)=0$$

Divide out common factors:

$$\sqrt{5-6x}\ln(2x-a)=0$$

Okay, now I would first look at the square root, and observe that in order for the equation to have only 1 solution, we need:

$$5-6x>0\implies x<\frac{5}{6}$$

Now, in order for the equation to be true with respect to the factor involving \(a\), we require:

$$2x-a=1\implies x=\frac{a+1}{2}$$

And so:

$$\frac{a+1}{2}<\frac{5}{6}$$

$$a+1<\frac{5}{3}$$

$$a<\frac{2}{3}$$

 

[Wilmer]

Okay, so the equation is:

$$\sqrt{5-6x}\ln\left(4x^2-a^2\right)=\sqrt{5-6x}\ln\left(2x+a\right)$$

Mark, seems to me that above easily/instantly simplifies to:
4x^2 - a^2 = 2x + a

No?

 

[Avalance789]

And after that simplification?

 

[MarkFL]

Mark, seems to me that above easily/instantly simplifies to:
4x^2 - a^2 = 2x + a

No?

Yes, and I saw that after going through the laborious log shuffle, but we will get to the same place. But, we still need to be mindful of the domain.

 

[Avalance789]

Just tell me if a<2/3 is correct answer. I will be happiest person on Earth, guys

 

[MarkFL]

Just tell me if a<2/3 is correct answer. I will be happiest person on Earth, guys

Something I didn't consider when I simplified is that:

$$4x^2-a^2>0$$

$$4\left(\frac{a+1}{2}\right)^2-a^2>0$$

$$(a+1)^2-a^2>0$$

$$(a+1+a)(a+1-a)>0$$

$$2a+1>0\implies a>-\frac{1}{2}$$

Now, suppose $$x=\frac{5}{6}$$

Then we have:

$$4\left(\frac{5}{6}\right)^2-a^2>0$$

$$\left(\frac{5}{3}\right)^2-a^2>0$$

$$|a|<\frac{5}{3}$$

So, what does all this imply?

 

[Greg]

4x^2 - a^2 = 2x + a

$$4x^2-a^2=(2x+a)(2x-a)$$

 

[Wilmer]

$$4x^2-a^2=(2x+a)(2x-a)$$

Yes...but this was the OP's:
$$\sqrt{5-6x}\ln\left(4x^2-a^2\right)=\sqrt{5-6x}\ln\left(2x+a\right)$$

 

[Greg]

Ah, pardon me.

 

[Wilmer]

Ah, pardon me.

Only if you recite the Hooooooly Rosary twice !

 

[Avalance789]

Ок, but if a=1, then x=1.

Cannot be that -5/3<a<5/3