Resolve Logarithmic inequation

[petarantes]

[MATH] \sqrt{log_a(\frac{3-2x}{1-x})}<1\\ solution: 1 < a < 2, 2 \leq x < \frac{a-3}{a-2}\\ a=2, x \geq 2\\ a >2, x < \frac{a-3}{a-2}\\ a > 2, x \geq 2\\ 0 < a < 1, \frac{a-3}{a-2}< x \leq2 \\ I ~try\\ D:\boxed{ \frac{3-2x}{1-x} > 0 \rightarrow x < 1 ~or ~x > \frac{3}{2}\\ and~log_a\frac{3-2x}{1-x}\geq0\rightarrow x\leq 1~or~x \geq \frac{3}{2},~if~a > 1\\ and ~log_a\frac{3-2x}{1-x} \leq 0 \rightarrow1 \leq x \leq\frac{3}{2}, if~ 0 < a < 1\\ and ~x\neq 1}\\ If~a > 1\rightarrow log_a\frac{3-2x}{1-x}< 1 \rightarrow x <\frac{a-3}{a-2}\\ If ~0 < a < 1 \rightarrow log_a \frac{3-2x}{1-x} > 1 \rightarrow x > \frac{a-3}{a-2}\\ \text{But I couldn't put the solutions together. Can someone help?} [/MATH]

 

[Steven G]

What are you trying to do? Are you trying to prove that log_a(3-2x)/(1-x) <1. If yes, you really should just state so.

 

[Dr.Peterson]

It's hard to be sure what your goal is; you need to use a few more words both to state the problem and to give the solution, as well as your work.

It looks like you want to solve the inequality for x, which depends on the value of the parameter a, so that the solution is stated in separate cases. But it seems odd that there are two cases both of which say a>2. Is this the solution given by a book? Is nothing more said about it?

I see that you are finding the domain of the LHS under two cases. Then in each of those cases you solved for x. I think your concern is that you can't get it into the form you were given. Is that right?

I haven't dug into it yet, but it looks to me like the next thing to do will be to look at how a affects each case, and break out more cases. For example, do you see why a can't be 2 under certain conditions?

EDIT: On glancing through the details, I see that the second and third lines in the D box were solved incorrectly, as if the log weren't there.

EDIT2: You may find it helpful to plot the solution on the ax-plane, as it will form something like three distinct regions.

 

[petarantes]

HI Dr. Peterson and Jomo.
The objective is clear, to solve the inequality, It is not to prove <1. It's in the title. It is a book issue.
The solution is from the wolfram website but I am putting the solution ((attachment) in the book which is the same as the website. (..."But it seems odd that there are two cases both of which say a>2 "...)
See that the book solution put the two cases together> 2.

 

[petarantes]

Dr. Peteson
EDIT: On glancing through the details, I see that the second and third lines in the D box were solved incorrectly, as if the log weren't there

Second line: [MATH]log_a \frac{3-2x}{1-x} \geq 0 \rightarrow \frac{3-2x}{1-x} \geq 1 \rightarrow x < 1~or~\geq 2 \rightarrow a > 1, x \neq 1[/MATH]
Third line: [MATH]log_a \frac{3-2x}{1-x} \geq 0 \rightarrow \frac{3-2x}{1-x} \leq 1 \rightarrow 1 < x \leq 2\rightarrow 0 < a < 1, x \neq 1 [/MATH]

 

[Dr.Peterson]

HI Dr. Peterson and Jomo.
The objective is clear, to solve the inequality, It is not to prove <1. It's in the title. It is a book issue.
The solution is from the wolfram website but I am putting the solution ((attachment) in the book which is the same as the website. (..."But it seems odd that there are two cases both of which say a>2 "...)
See that the book solution put the two cases together> 2.

That form is considerably clearer, connecting them with "or". It troubles me when I see students list inequalities without indicating how they are to be combined. I suppose I can forgive software for doing that.

Dr. Peteson
EDIT: On glancing through the details, I see that the second and third lines in the D box were solved incorrectly, as if the log weren't there

Second line: [MATH]log_a \frac{3-2x}{1-x} \geq 0 \rightarrow \frac{3-2x}{1-x} \geq 1 \rightarrow x < 1~or~\geq 2 \rightarrow a > 1, x \neq 1[/MATH]
Third line: [MATH]log_a \frac{3-2x}{1-x} \geq 0 \rightarrow \frac{3-2x}{1-x} \leq 1 \rightarrow 1 < x \leq 2\rightarrow 0 < a < 1, x \neq 1 [/MATH]

That's more like it.

Now, I think you need to combine the solution in each case (at the end of your work) with the domain information. This is where graphing each region may help make sense of how everything fits together.

For example, in the case [MATH]0<a<1[/MATH], you have [MATH]x>\frac{a-3}{a-2}[/MATH] and [MATH]1\lt x\le 2[/MATH]. What does this region look like? Which fact can be dropped? What line in the given solution does this correspond to?

The most interesting part will involve the asymptote at [MATH]a=2[/MATH].

 

[petarantes]

As no one was able to solve it I will post the solution I found. It is not an easy exercise.

[MATH]\sqrt{log_a\frac{3-2x}{1-x}}<1 \Leftrightarrow 0 \leq log_a \frac{3-2x}{1-x}< 1\\ \boxed{Step1: 0 < a < 1}\\ 0 \leq log_a \frac{3-2x}{1-x}< 1\\ \begin{cases} \frac{3-2x}{1-x} > a \rightarrow \frac{(-2+a)x+3-a}{1-x} > 0(I)\\ \frac{3-2x}{1-x}\leq 1 \rightarrow\frac{-x+2}{1-x} \leq0(II) \end{cases}\\I: If~0 < a < 1 \rightarrow-2+a < 0 \rightarrow decreasing~ function\\ I:Solving: x < 1~or~ x > \frac{a-3}{a-2}\\ II:Solving: 1 < x \leq 2\\ I\cap II: \frac{a-3}{a-2} < x < 2 \rightarrow \\ \boxed{\color{red}0 < a < 1 \Rightarrow S_1 = \{x\in \mathbb{R} | \ \frac{a-3}{a-2} < x < 2 \}}[/MATH]

[MATH]\boxed{Step2: a > 1~and~-2+a < 0 \rightarrow 1 < a < 2} \\ 0 \leq log_a \frac{3-2x}{1-x}< 1 \Rightarrow 1 \leq \frac{3-2x}{1-x}< a \\ \begin{cases} \frac{3-2x}{1-x} \geq 1 \rightarrow \frac{-x+2}{1-x} \geq 0(I)\\ \frac{3-2x}{1-x} < a \rightarrow\frac{(-2+a)x+3-a}{1-x} < 0(II) \end{cases}\\II: If~ a > 1 \rightarrow-2+a < 0 \rightarrow decreasing~ function\\ I:Solving: x < 1~or~ x > \frac{a-3}{a-2}\\ II:Solving: x < 1 ~or ~x \geq 2\\ I\cap II: 2 \leq x < \frac{a-3}{a-2} \rightarrow \\ \boxed{\color{red}1 < a < 2 \Rightarrow S_2 = \{x\in \mathbb{R} | \ 2 \leq\ x \leq \frac{a-3}{a-2} \}}[/MATH]
[MATH] \boxed{Step3: a > 1~and~-2+a = 0 \rightarrow a = 2} \\ 0 \leq log_a \frac{3-2x}{1-x}< 1 \Rightarrow 1 \leq \frac{3-2x}{1-x}< 2 \\ \begin{cases} \frac{3-2x}{1-x} \geq 1 \rightarrow \frac{-x+2}{1-x} \geq 0(I)\\ \frac{3-2x}{1-x} < 2 \rightarrow\frac{1}{1-x} < 0(II) \end{cases}\\ I:Solving: x < 1~or~ x \geq 2\\ II:Solving: x > 1\\ I\cap II: x \geq 2 \rightarrow \\ \boxed{\color{red} a=2 \Rightarrow S_3 = \{x\in \mathbb{R} | x \geq 2 \}}[/MATH]
[MATH]\boxed{Step4: a > 1~and~-2+a > 0 \rightarrow a > 2} \\ 0 \leq log_a \frac{3-2x}{1-x}< 1 \Rightarrow 1 \leq \frac{3-2x}{1-x}< a \\ \begin{cases} \frac{3-2x}{1-x} \geq 1 \rightarrow \frac{-x+2}{1-x} \geq 0(I)\\ \frac{3-2x}{1-x} < a \rightarrow\frac{(-2+a)x+3-a}{1-x} < 0(II) \end{cases}\\II: If~ a > 2 \rightarrow-2+a < 0 \rightarrow increasing~ function\\ I:Solving: x < 1~or~ x \geq 2\\ II:Solving: x < \frac{a-3}{a-2} ~or ~x >1\\ I\cap II: x < \frac{a-3}{a-2} ~or ~x \geq 2 \rightarrow \\ \boxed{\color{red}a > 2 \Rightarrow S_4 = \{x\in \mathbb{R} | x < \frac{a-3}{a-2} ~or ~x \geq 2 \}}[/MATH]
[MATH]\boxed{\color{red}\mathsf{S=S_1\cup S_2\cup S_3\cup S_4}}[/MATH]

 

[Dr.Peterson]

As no one was able to solve it I will post the solution I found. It is not an easy exercise.

Who said no one was able to solve it? I was trying to help you do so, and had no intention of writing up my work as a complete solution, because that's not my goal here. Did you follow my suggestions?

But you're right, it isn't easy. Where did it come from, and why were you hoping to solve it yourself?