<s>[INST] 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!
 [/INST] 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:
$$x=\left(\frac{a}{(a+1)^2}\right)^{\frac{1}{2\log_{b}a-1}}$$
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