Logarithms 101

You may use the fact that $\log_a b = \cfrac{1}{\log_b a}$

$4 \log_2 x = 4 - \log_x 2 \\ \Rightarrow 4 U = 4 - \cfrac{1}{U} \\ \Rightarrow 4 U^2 = 4U - 1 \\ \Rightarrow 4 U^2 - 4U + 1 = 0 \\ \Rightarrow (2U-1)^2 = 0 \\ \Rightarrow U = \cfrac{1}{2} \\ \Rightarrow \log_2 x = U = \cfrac{1}{2} \Rightarrow \log_2 x = \log_2 2^{ \frac{1}{2} } \\ \Rightarrow x = \sqrt{2}$

$\begin{aligned} 4\log_2 x & = 4 - \log_x 2 & \small \blue{\text{Converting to the common log base}} \\ \frac {4 \log x}{\log 2} & = 4 - \frac {\log 2}{\log x} & \small \blue{\text{As } x>1 \text{, multiply both sides by }\log 2 \log x} \\ 4 \log^2 x & = 4 \log 2 \log x - \log^2 2 \\ 4 \log^2 x - 4 \log 2 \log x + \log^2 2 & = 0 \\ \left(2\log x - \log 2\right)^2 & = 0 \\ \implies \log x^2 & = \log 2 \\ x^2 & = 2 & \small \blue{\text{Note that }x > 1} \\ x & = \sqrt 2 \approx \boxed{1.41} \end{aligned}$

Using logarithm properties, we can switch the base of one of these logs (I chose the one on the left, but switching the other will lead to the same conclusion): $4\log_{2}x = 4 - \log_{x}2 \Rightarrow \frac{4}{\log_{x}2} = 4 - \log_{x}2 \; \; \; \; \left(\text{Property: } \log_{a}b \equiv \frac{1}{\log_{b}a}\right)$

We can tidy things up a bit by multiplying by $\log_{x}2$ on both sides: $\frac{4}{\log_{x}2} = 4 - \log_{x}2 \Rightarrow 4 = 4\log_{x}2 - (\log_{x}2)^2 \Rightarrow (\log_{x}2)^2-4\log_{x}2+4 = 0$

This looks like a quadratic equation! We can factorise this and solve for $\log_{x}2$ before solving for $x$ itself: $(\log_{x}2)^2-4\log_{x}2+4 = 0 \Rightarrow (\log_{x}2 - 2))^2 = 0 \Rightarrow \log_{x}2 = \boxed{2}$

Now that we know what $\log_{x}2$ is, solving for $x$ is easy: $\huge \log_{x}2 = 2 \Rightarrow x = \boxed{\sqrt{2}}$

I don't know the relevant properties of logarithms, so I replaced some numbers.

Let $a=\log_2x$ and $b=\log_x2$ , so we can get $x=2^a$ and $2=x^b$ .

Substitute $x=2^a$ into $2=x^b$ to get $2=2^{ab}$ , so $ab=1$ .

Then change the title to $4a=4-b$ , and solve the system of equations to get $b=2$ , so

$x=\sqrt{2}\approx\boxed{1.41}$ .

We can rewrite as follows: $4a=4-\dfrac{1}{a}$ $4a+\dfrac{1}{a}=4$ $4a^2+1=4a$ $4a^2-4a+1=0$ $\implies a=\dfrac{1}{2}$ $\implies \log_{2}(x)=\dfrac{1}{2}$ $\boxed{x=\sqrt{2}}$