Parametric equations - is my answer right. 
Find all values of $a$ for which the equation
  $$ (a-1)4^x + (2a-3)6^x = (3a-4)9^x $$
  has only one solution.


I have two cases, one when $a = 1$ and other when Discriminant $= 0$.
The answers I get are $a =1; a=0,5$. 
Since I don't have the answers could sb tell me if I am right.
 A: Write $t=3^x/2^x>0$, then we have $$(a-1)+(2a-3)t=(3a-4)t^2$$
Case $a={4\over 3}$ then the equation is linear so $$t= {1-a\over 2a-3}=1\implies \Big({3\over 2}\Big)^x= 1 \implies x=0$$
Case $a\ne {4\over 3}$ then the equation is quadratic, so we have two subcases 


*

*The discirminat is $0$: $$(2a-3)^2+4(3a-4)(a-1)=0$$ which is the same 
as $$(4a-5)^2 =0$$ so $a={5\over 4}$. In this case we get $t=1$ again.

*The discirminat is $>0$ then we have two solutions $t_1,t_2$ and if we want to have exactly one then one must be positive and other negative, so $$t_1\cdot t_2 <0\implies {1-a\over 3a-4}<0 \implies a\in (-\infty, 1]\cup [{4\over 3},\infty)\cup\{{5\over 4}\}$$
(Last inequality we get from Vieta formulas.)

A: Here is another way to solve this. 
Let $t=2^x/3^x>0$, then we get $$a(t^2+2t-3) = t^2+3t-4$$ so $$a(t+3)(t-1)=(t+4)(t-1)$$
Then for each $a$ number $t=1$ i.e. $x=0$ is a solution. Say $t\neq 1$, then $$a ={t+4\over t+3}\implies a\in (1,{4\over 3})$$
So for each $a\in (1,{4\over 3})$ we have another solution, so $$\boxed{a\in (-\infty, 1]\cup [{4\over 3},\infty)}$$

Example, if $a=2$ we get $$2\cdot 9^x = 4^x+6^x\implies t\in\{1,-2\}$$
But $t>0$ so we have only 1 solution. 
