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$\sqrt{log_a(\frac{3-2x}{1-x})}<1$ wolfram's 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}$

But I couldn't put the solutions together. Can someone help?

peta arantes
  • 6,211
  • $\log_a [(3-2x)/(1-x)] \geq 0$ and $a>1$ should not give the same answer as $(3-2x)/(1-x) \geq 0$. And $\log_a [(3-2x)/(1-x)] < 0$ means $x$ is not in the domain at all, since its square root will be taken (and must be real). – aschepler Apr 16 '20 at 02:55
  • Forget Wolfram Alpha results : in this case they are too confusing : restart from scratch by yourself... First step : graph the $log_a$ and $exp_a$ functions : you will see they are increasing (general case) or decreasing (exceptional case when $a<1$ that you can drop in a first approach). Then $\sqrt{A}<B \iff A<B^2 \iff exp_a(A)<exp_a(B^2)$ etc... – Jean Marie Apr 16 '20 at 06:39
  • Thanks Jean Marie, but I cant use graph, i need a algebric solution. – peta arantes Apr 16 '20 at 13:59

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