# 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:

$$(1)\Leftrightarrow 3^{\log_{4}x}.\sqrt{3}+ \frac{3^{\log_{4}x}}{\sqrt{3}} = \sqrt{x}$$ $$\Leftrightarrow 3^{\log_{4}x}.3+ 3^{\log_{4}x} = \sqrt{3x}$$ $$\Leftrightarrow 4.3^{\log_{4}x}= \sqrt{3x}$$

First of all impose the necessary existence conditions, that is: $$x > 0$$ for the logarithms and $$x \geq 0$$ for the square root. That is, eventually,

$$x > 0$$

for the whole equation.

Then follow Siong Thye Goh reasoning, obtaining the final equation he wrote.

At that point:

$$1 + \log_4(3)\log_4(x) = \frac{\log_4(3)}{2} + \frac{\log_4(x)}{2}$$

$$\log_4(x)\left(\log_4(3) - \frac{1}{2}\right) = \frac{\log_4(3)}{2} - 1$$

$$\log_4(x) = \frac{\frac{\log_4(3)}{2} - 1}{\log_4(3) - \frac{1}{2}} = \frac{\log_4(3)-2}{2\log_4(3)-1}$$

To solve for $$x$$ take the exponential base 4 of both terms, getting:

$$\large x = \large 4^{\frac{\log_4(3)-2}{2\log_4(3)-1}}$$

Guide:

Taking $$\log_4$$ on both sides,

$$1 + \log_4 3 \cdot \log_4 x = \frac12 (\log_43 + \log_4 x)$$

Solve for $$\log_4 x$$.

You may also continue as follows:

$$\begin{eqnarray*} 4 \cdot 3^{2 \cdot \log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\ 4 \cdot 9^{\log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\ 4 \cdot 4^{\log_4{9} \cdot \log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\ (\sqrt{x})^{\log_4{9} -1} &= & \frac{\sqrt{3}}{4} \Leftrightarrow\\ x & = & \left( \frac{\sqrt{3}}{4}\right)^{\frac{\log_4{9}-1}{2}} \approx 0.0571725 \end{eqnarray*}$$

Let's generalise a bit, with parameters say $$a, b \in (0, \infty)\setminus \{1\}$$ subject to $$a^2 \neq b$$ and let's try to solve the equation:

$$a^{\log_{b}x+\frac{1}{2}}+a^{\log_{b}x-\frac{1}{2}}=\sqrt{x}$$

Notice that the left-hand side can be rewritten as

$$a^{\log_{b}x}(\sqrt{a}+\frac{1}{\sqrt{a}})=b^{\log_{b}a\cdot \log_{b}x}(\sqrt{a}+\frac{1}{\sqrt{a}})=x^{\log_{b}a}(\sqrt{a}+\frac{1}{\sqrt{a}})$$

As you are dealing exclusively with strictly positive reals, your given equation is equivalent to its square, so to speak:

$$\frac{(a+1)^2}{a}x^{2\log_{b}a}=x$$

$$x^{2\log_{b}a-1}=\frac{a}{(a+1)^2}$$
Since the right-hand side is never $$1$$ (you can try to see why), this is why we initially imposed the relation of inequality between $$a$$ and $$b$$; it is satisfied in the particular case of your equation.
$$x=\left(\frac{a}{(a+1)^2}\right)^{\frac{1}{2\log_{b}a-1}}$$