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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 at 10:28
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why did you remove your work? Leave it, it shows your effort... –  draks ... Jul 15 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 at 10:31
    
come on you can even do that in a minute...I believe in you... –  draks ... Jul 15 at 10:32
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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 at 10:33

2 Answers 2

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

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

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

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

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