<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] $$
3^{log_4x+\frac{1}{2}}+3^{log_4x-\frac{1}{2}}=\sqrt{x}\\
(3^{\frac{1}{2}}+3^{-\frac{1}{2}}) 3^{log_4x}=\sqrt{x}\\
$$
Write $3 = 4^{\log_4 3}$:
$$
(3^{\frac{1}{2}}+3^{-\frac{1}{2}}) 4^{\log_4 (3) \cdot log_4x}=\sqrt{x}\\
$$
Take log to base 4:
$$
\log_4(3^{\frac{1}{2}}+3^{-\frac{1}{2}}) +\log_4 (3) \cdot \log_4x=\frac12 \log_4 x\\
$$
So
$$
\log_4(x) = \frac{\log_4(3^{\frac{1}{2}}+3^{-\frac{1}{2}})}{\frac12 -\log_4 (3) }
$$
and
$$
x = 4^{\frac{\log_4(3^{\frac{1}{2}}+3^{-\frac{1}{2}})}{\frac12 -\log_4 (3) }}
$$
or, simplified further, 
$$
x = 4^{\frac{2\log_4(3^{\frac{1}{2}}+3^{-\frac{1}{2}})}{1-2\log_4 (3) }}\\
= 4^{\frac{\log_4((3^{\frac{1}{2}}+3^{-\frac{1}{2}})^2)}{1-2\log_4 (3) }} = 
4^{\frac{\log_4(3 + 2 + \frac13)}{1-2\log_4 (3) }} \\
 = 
4^{\frac{\log_4(\frac{16}{3})}{1-2\log_4 (3) }} =(\frac{16}{3})^ {\frac{1}{1-2\log_4 (3) }}
$$
You may also want to  write it with the $\exp$ function or find some other convenient way of expressing it.
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