Solving this equation: $3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}$ 
Solve this equation: 
  $$3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}\qquad  (1)$$

I tried to make both sides of the equation have a same base and I started:
$$(1)\Leftrightarrow 3^{\log_{4}x}.\sqrt{3}+ \frac{3^{\log_{4}x}}{\sqrt{3}} = \sqrt{x}$$
$$\Leftrightarrow 3^{\log_{4}x}.3+ 3^{\log_{4}x}   = \sqrt{3x}$$
$$\Leftrightarrow 4.3^{\log_{4}x}= \sqrt{3x}$$
At this step, I can't continue. Please help me!

First of all impose the necessary existence conditions, that is: $x > 0$ for the logarithms and $x \geq 0$ for the square root.
That is, eventually, 
$$x > 0$$
for the whole equation. 
Then follow Siong Thye Goh reasoning, obtaining the final equation he wrote.
At that point:
$$1 + \log_4(3)\log_4(x) = \frac{\log_4(3)}{2} + \frac{\log_4(x)}{2}$$
$$\log_4(x)\left(\log_4(3) - \frac{1}{2}\right) = \frac{\log_4(3)}{2} - 1$$
$$\log_4(x) = \frac{\frac{\log_4(3)}{2} - 1}{\log_4(3) - \frac{1}{2}} = \frac{\log_4(3)-2}{2\log_4(3)-1}$$
To solve for $x$ take the exponential base 4 of both terms, getting:
$$\large x =  \large 4^{\frac{\log_4(3)-2}{2\log_4(3)-1}}$$
