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?

If you let $x=u+{1\over2}$, you can get rid of the pesky square roots: the expression simplifies to
$$9^{u+1/2}-2^{u+1}=2^{u+4}-3^{2u}$$
or
$$3\cdot9^u-2\cdot2^u=16\cdot2^u-9^u$$
This simplifies first to $4\cdot9^u=18\cdot2^u$ and then to $9^{u-1}=2^{u-1}$, which clearly implies $u=1$, i.e., $x=3/2$.
