<s>[INST] Solve $9^x-2^{x+\frac{1}{2}}=2^{x+\frac{7}{2}}-3^{2x-1}$ $$9^x-2^{x+\frac{1}{2}}=2^{x+\frac{7}{2}}-3^{2x-1}.$$
The equation states solve for $x$.
What I first did was put like bases together.
$$3^{2x}+3^{2x-1}= 2^{x+\frac{7}{2}}+ 2^{x+\frac{1}{2}}.$$
Then I factored $3^{2x}$ and $2^x$
$$3^{2x}(1+\frac{1}{3})=2^x(2^{\frac{7}{2}}+2^{\frac{1}{2}}),$$
then I got
$$\frac{3^{2x}}{2^x}=9\sqrt{2}.$$
From here I took $\log$s, but the answer wasn't nice. 
What to do?
 [/INST] $$9^x-2^{x+\frac12}=2^{x+\frac72}-3^{2x-1}$$
$$\to\frac43(9^x)=9(2^{x+\frac12})$$
Hence we form the two iterates:
$$x_{n+1}=\log_9{\bigg[\frac{27}{4}(2^{x_n+\frac12})\bigg]}$$
and
$$x_{n+1}=\log_2{\bigg[\frac{4}{27}(9^{x_n})\bigg]}$$
The first gives the solution $x=\frac32$, the second diverges to $-\infty$, and thus there is no second solution
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