<s>[INST] 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?
 [/INST] We have
$$
\begin{eqnarray}
  \sqrt{x}
  &=& 3^{\color{blue}{\log_4(x)} \color{brown}{+ \frac{1}{2}}} + 3^{\color{blue}{\log_4(x)} \color{brown}{- \frac{1}{2}}} \\
  &=& \left( \color{brown}{3^{\frac{1}{2}}} + \color{brown}{3^{-\frac{1}{2}}} \right) \cdot \color{blue}{3^{\log_4(x)}} \tag{factoring out $3^{\log_4(x)}$} \\
  &=& \left( \color{brown}{3^{\frac{1}{2}}} + \color{brown}{3^{-\frac{1}{2}}} \right) \cdot \color{blue}{x^{\log_4(3)}} \tag{using $a^{\log_b(c)} = c^{\log_b(a)}$} \\
  &=& \left( \color{brown}{\sqrt{3}} + \color{brown}{\dfrac{1}{\sqrt{3}}} \right) \cdot \color{blue}{x^{\log_4(3)}} \tag{using $a^\frac{1}{2} = \sqrt{a}$}
\end{eqnarray}
$$
If we square both sides, we get
$$
x
= \left( \color{brown}{\sqrt{3}} + \color{brown}{\dfrac{1}{\sqrt{3}}} \right)^2 \cdot \color{blue}{x}^{2\color{blue}{\log_4(3)}}
= \dfrac{16}{3} \cdot x^{\log_4(9)} \;.
$$
Since $x \neq 0$, we are allowed to divide by $x^{\log_4(9)}$. If we do this, we obtain
$$
\dfrac{16}{3} = x^{1 - \log_4(9)} = x^{\log_4\left( \frac{4}{9} \right)} \;.
$$
Now, we raise both sides to the power of $\log_{\frac{4}{9}}(4)$ and, using the identity $\log_{4}\left( \frac{4}{9} \right) \cdot \log_{\frac{4}{9}}(4) = 1$, obtain
$$
x
= \left( \dfrac{16}{3} \right)^{\log_{\frac{4}{9}}(4)}
\approx 0.0572 \;.
$$
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