# Solving ln/exponent question

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

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

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Exponentiate rhs and lhs, may be ! – Claude Leibovici Jul 15 '14 at 10:28
why did you remove your work? Leave it, it shows your effort... – draks ... Jul 15 '14 at 10:29
Because when I did my own work I did not notice there was a - (minus sign) which makes everything completely wrong – ADGB Jul 15 '14 at 10:31
come on you can even do that in a minute...I believe in you... – draks ... Jul 15 '14 at 10:32
I think it is y^1/2=e^(x^(1/2)-4.105) but then there should be two results, a positive and a negative shouldn't it? – ADGB Jul 15 '14 at 10:33

HINT: What is the inverse function to $\ln$?
$$\exp(\ln(f(x)))=f(x)$$
$$x=[4.105-\ln(\sqrt{y})]^2 \Rightarrow \pm\sqrt{x}=4.105-\ln(\sqrt{y}) \Rightarrow \pm\sqrt{x}=4.105-\ln({y}^{\frac{1}{2}}) \Rightarrow \\ \pm\sqrt{x}=4.105-\frac{1}{2}\ln({y}) \Rightarrow \ln{(y)}=2 \cdot 4.105\pm2 \sqrt{x} \Rightarrow y=e^{2 \cdot 4.105\pm2 \sqrt{x}}$$