Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


share|cite|improve this question
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

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

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


share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.