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!
 A: Guide:
Taking $\log_4$ on both sides,
$$1 + \log_4 3 \cdot \log_4 x = \frac12 (\log_43 + \log_4 x)$$
Solve for $\log_4 x$.
A: 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:
$$\large x =  \large 4^{\frac{\log_4(3)-2}{2\log_4(3)-1}}$$
A: Same solution with same approach 
But simplified with substitution 

A: You may also continue as follows:
$$\begin{eqnarray*}
4 \cdot 3^{2 \cdot \log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\
4 \cdot 9^{\log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\
4 \cdot 4^{\log_4{9} \cdot \log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\
(\sqrt{x})^{\log_4{9} -1} &= & \frac{\sqrt{3}}{4} \Leftrightarrow\\
x & = & \left( \frac{\sqrt{3}}{4}\right)^{\frac{2}{\log_4{9}-1}} \approx 0.0571725
\end{eqnarray*}$$
A: Let's generalise a bit, with parameters say $a, b \in (0, \infty)\setminus \{1\}$ subject to $a^2 \neq b$ and let's try to solve the equation:
$$a^{\log_{b}x+\frac{1}{2}}+a^{\log_{b}x-\frac{1}{2}}=\sqrt{x}$$
Notice that the left-hand side can be rewritten as 
$$a^{\log_{b}x}(\sqrt{a}+\frac{1}{\sqrt{a}})=b^{\log_{b}a\cdot \log_{b}x}(\sqrt{a}+\frac{1}{\sqrt{a}})=x^{\log_{b}a}(\sqrt{a}+\frac{1}{\sqrt{a}})$$
As you are dealing exclusively with strictly positive reals, your given equation is equivalent to its square, so to speak:
$$\frac{(a+1)^2}{a}x^{2\log_{b}a}=x$$
which leads to
$$x^{2\log_{b}a-1}=\frac{a}{(a+1)^2}$$
Since the right-hand side is never $1$ (you can try to see why), this is why we initially imposed the relation of inequality between $a$ and $b$; it is satisfied in the particular case of your equation.
We finally have the solution:
$$x=\left(\frac{a}{(a+1)^2}\right)^{\frac{1}{2\log_{b}a-1}}$$
