<s>[INST] For all real $x$ the expression$ \frac{x^2-2x+4}{x^2+2x+4}$ lies between $\frac 13$ and $3$. 
The values between which the expression $$\frac{9.3^{2x}+6.3^x+4}{9.3^{2x}-6.3^x+4}$$ is

A solution I saw says that the expression can be written as 
$$\frac{(3.3^x)^2+2(3.3^x)+4}{(3.3^x)^2-2.(3.3^x)+4}$$ which is analogous to the previous expression and hence would have the same interval. Why are they analogous? Clearly it has $x$ as a power, so something has to change right? Why are they the same?
 [/INST] $$\frac{9.3^{2x}+6.3^x+4}{9.3^{2x}-6.3^x+4}$$
$$\frac{(3)^2(3^x)^2+2(3)(3^x)+4}{(3)^2(3^x)^2-2(3)(3^x)+4}$$
$$\frac{(3.3^x)^2+2(3.3^x)+4}{(3.3^x)^2-2(3.3^x)+4}$$
$$\frac{(z)^2+2(z)+4}{(z)^2-2(z)+4}$$
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