If $9^{x-\frac{1}{2}} - 2^{2x-2} = 4^x - 3^{2x-3}$, then x is

Given: $9^{x-\frac{1}{2}} - 2^{2x-2} = 4^x - 3^{2x-3}$

For complex exponential equations like this, testing values is usually faster than full algebraic manipulation.

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Let's test $x = \frac{3}{2}$:

Left side calculation:

$9^{x-\frac{1}{2}} = 9^{\frac{3}{2}-\frac{1}{2}} = 9^1 = 9$

$2^{2x-2} = 2^{2(\frac{3}{2})-2} = 2^{3-2} = 2^1 = 2$

So left side $= 9 - 2 = 7$

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Right side calculation:

$4^x = 4^{\frac{3}{2}} = (2^2)^{\frac{3}{2}} = 2^3 = 8$

$3^{2x-3} = 3^{2(\frac{3}{2})-3} = 3^{3-3} = 3^0 = 1$

So right side $= 8 - 1 = 7$

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Since both sides equal $7$, our solution is verified.

Remember: $a^{m-n} = \frac{a^m}{a^n}$ and $a^0 = 1$ for any non-zero base $a$.

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Therefore: $x = \frac{3}{2}$