solve for x?

Thank you so much guys!

$$log_4(4) = 1$$

$$\\\log_4(x)=\sqrt{\log_4(x)} \\ \\ \quad x=4 \Rightarrow \log_4(4)=\sqrt{\log_4(4)}\Rightarrow 1 = \sqrt1 = 1$$

$$\boxed{x=4}$$

...or

$$\\\log_4(x)=\sqrt{\log_4(x)} \quad | \quad 1^2\\ \\ \log_4(x)\times\log_4(x)=\log_4(x) \quad | \quad :\log_4(x)\\ \\ \log_4(x) = 1\quad | \quad 4^x\\ \\ 4^{\log_4(x)} =x= 4^1 \\ \boxed{x=4}$$

$$log_4(x)=\sqrt{log_4(x)}$$

$$(log_4(x))^2 = log_4(x)$$

$$(log_4(x))^2-log_4(x) = 0$$

$$log_4(x)(log_4(x)-1) = 0$$

$$log_4(x) = 0$$  or  $$log_4(x) = 1$$

$$x = 1$$  or  $$x = 4^1 = 4$$

Sorry for the double solution, but I wanted a variant which also gave  $$x=1$$

Reinout