Find the solution of the equation $9^x - 2^{x + 1/2} = 2^{x + 3/2} - 3^{2x - 1}$.

$0.5737373...... = $

If ${a^x} = {(x + y + z)^y}$,${a^y} = {(x + y + z)^z}$,and ${a^z} = {(x + y + z)^x}$,then:

If $x = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$ and $y = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$,then find the value of $3x^2 + 4xy - 3y^2$.

Let $p = 99$ and $q = 101$. Define $p_1 = \log_{10} \left(\frac{p+q}{2}\right)$ and $q_1 = \frac{1}{2}(\log_{10} p + \log_{10} q)$,and $p_2 = \log_{10} \left(\frac{p_1+q_1}{2}\right)$,$q_2 = \frac{1}{2}(\log_{10} p_1 + \log_{10} q_1)$. Then:

The value of $a^{\log_b x}$,where $a = 0.2$,$b = \sqrt{5}$,and $x = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$ to $\infty$ is: