$\ \log _x\left(\log _{36}\left(2\cdot 9^{2x}-3\cdot 4^{2x}\right)\right)<\:1 $ I have to find x in the inequation below:
$$\ \log _x\left(\log _{36}\left(2\cdot 9^{2x}-3\cdot 4^{2x}\right)\right)<\:1 $$
So I did the following:
$$\ \log _{36}\left(2\cdot \:9^{2x}-3\cdot \:4^{2x}\right)<\:x $$
$$\ 2\cdot \:9^{2x}-3\cdot \:4^{2x}<\:36^x $$
$$\ 2-3\cdot \left(\frac{4}{9}\right)^{2x}<\left(\frac{4}{9}\right)^x $$
$$\ u=\left(\frac{4}{9}\right)^x $$
$$\ 3u^2+u-2>0 $$
Which results in $\ \left(\frac{4}{9}\right)^x=-1 $ and
$\ \left(\frac{4}{9}\right)^x=\frac{2}{3} $ so from the first one x isn't real, and from the second $$\ x=\frac {1}{2} $$ 
Also verifing that $\ 2\cdot 9^{\left(2x\right)}-3\cdot 4^{\left(2x\right)}>0 $ gives $\ x> \frac{1}{4} $. With the correct answer being $\ \frac{1}{2}<x<1 $. Could you let me know what am I missing and if my approach is good?
 A: You have to consider two different cases separately: $0<x<1$ and $x>1$. The solution you presented here is only valid for the latter case when $x>1$. But in the case $0<x<1$, logarithm base $x$ is a decreasing function, and thus eliminating it would flip the sign of the inequality — so that your second line will be $\log_{36}(\cdots)\color{red}{>}x$.
By the way, you seem to have made a similar mistake when finishing your solution too. Solving the quadratic inequality gives you $u<-1$ or $u>\frac{2}{3}$. As you pointed out, the equation $\left(\frac{4}{9}\right)^x=-1$ has no solutions; moreover, the inequality $\left(\frac{4}{9}\right)^x<-1$ has no solutions either. But the other root gives you an inequality
$$u>\frac{2}{3} \quad \Longrightarrow \quad \left(\frac{4}{9}\right)^x>\frac{2}{3} \quad \Longrightarrow \quad x\color{red}{<}\frac{1}{2}.$$
Again, the inequality sign flips because the base is less than $1$. Notice that this solution was for the case $x>1$, so in the end we see that this case has no solutions (the outcome $x<\frac{1}{2}$ contradicts the case condition $x>1$). But you still have the other case to solve.
