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?
 A: $$
3^{log_4x+\frac{1}{2}}+3^{log_4x-\frac{1}{2}}=\sqrt{x}\\
(3^{\frac{1}{2}}+3^{-\frac{1}{2}}) 3^{log_4x}=\sqrt{x}\\
$$
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.
A: 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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A: Use change of base formula:
$$\log_4 x = \frac{\log_3 x}{\log_3 4}$$
Can then get (with $p = \frac{1}{\log_3 4}$)
$$\sqrt{x} = 3^{\frac{1}{2}} (3^{\log_3 x})^p + 3^{-\frac{1}{2}} (3^{\log_3 x})^p = 3^{\frac{1}{2}} x^p + 3^{-\frac{1}{2}} x^{p} = x^p (\sqrt{3} + \frac{1}{\sqrt{3}})$$
So equation to solve becomes (with $a = (\sqrt{3} + \frac{1}{\sqrt{3}})$ )
$$\sqrt{x} = a x^p$$
So $x = 0$ is one obvious solution (which checks out in the original equation, since $\lim_{x \to 0} \log_b x = -\infty$ and $3^{-\infty}=0$). Otherwise,
$$ax^{p-\frac{1}{2}} = 1$$
so then just solve that with our constant values for $a$ and $p$. There's one real solution (on top of $0$), I think there are no complex solutions as this involves irrational exponents. (Haven't done this sort of thing in a while so not sure).
A: 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
$$
x
= \left( \dfrac{16}{3} \right)^{\log_{\frac{4}{9}}(4)}
\approx 0.0572 \;.
$$
