Solve the equation $3^{\log_4(x)+\frac{1}{2}}+3^{\log_4(x)-\frac{1}{2}}=\sqrt{x}$ 
Solve $3^{\log_4(x)+\frac{1}{2}}+3^{\log_4(x)-\frac{1}{2}}=\sqrt{x}$.

I am able to reduce the LHS to $\sqrt{x}=3^{\log_4(x)} \cdot \dfrac{4}{3}$. Squaring both sides do not seem to lead to a result. Do you know how to proceed?

Easy step by step
$$\begin{align}
3^{\log_4x+\frac{1}{2}}+3^{\log_4x-\frac{1}{2}}&=\sqrt{x} \\
\sqrt{3}\cdot 3^{\log_4x} + (\sqrt 3)^{-1} \cdot 3^{\log_4x}&=\sqrt x \\
3 \cdot 3^{\log_4x}+3^{\log_4x}&=\sqrt{3x} \\
4\cdot3^{\log_4x}&=\sqrt{3x} \\
3^{\log_4x}&=\frac{\sqrt{3x}}{4} \\
4^{(\log_4 3) \cdot (\log_4 x)} &=\frac{\sqrt{3x}}{4} \\
(\color{red}{4^{\log_4 x}})^{\log_4 3} &= \frac{\sqrt{3x}}{4} \quad \text{a logarithm in an exponent is an inverse operation}\\
x^{\log_4 3}&=\frac{\sqrt 3}{4} \cdot \sqrt{x} \\
x^{2\log_4 3}&=\frac{3}{16} \cdot x \\
&\color{red}{x \neq 0} \quad \text{by the original equation} \\
x^{2\log_4 3 -1}&=\frac{3}{16} \\
x&=\left(\frac{3}{16}\right)^{\frac{1}{{2\log_4 3 -1}}} \approx 0.0571725372071
\end{align}$$
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