Solving ln/exponent question

Question

How do I change the subject of the equation from x to y in the following equation:

$$x=[4.105-\ln(\sqrt{y})]^2$$

Answer 1

$$x=[4.105-\ln(\sqrt{y})]^2 \Rightarrow \pm\sqrt{x}=4.105-\ln(\sqrt{y}) \Rightarrow \pm\sqrt{x}=4.105-\ln({y}^{\frac{1}{2}}) \Rightarrow \\ \pm\sqrt{x}=4.105-\frac{1}{2}\ln({y}) \Rightarrow \ln{(y)}=2 \cdot 4.105\pm2 \sqrt{x} \Rightarrow y=e^{2 \cdot 4.105\pm2 \sqrt{x}}$$

Answer 2

Since this is where you got stuck previously, here is my

HINT: What is the inverse function to $\ln$?

$$\exp(\ln(f(x)))=f(x)$$