<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] You may also continue as follows:
$$\begin{eqnarray*}
4 \cdot 3^{2 \cdot \log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\
4 \cdot 9^{\log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\
4 \cdot 4^{\log_4{9} \cdot \log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\
(\sqrt{x})^{\log_4{9} -1} &= & \frac{\sqrt{3}}{4} \Leftrightarrow\\
x & = & \left( \frac{\sqrt{3}}{4}\right)^{\frac{2}{\log_4{9}-1}} \approx 0.0571725
\end{eqnarray*}$$
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