The discussion focuses on solving the equation \( y = \exp\left(\frac{-x\pi}{\sqrt{1-x^2}}\right) \) for \( x \) when \( y = 0.1 \). The solution involves using the natural logarithm, leading to the equation \( \ln(0.1) = \frac{-x\pi}{\sqrt{1-x^2}} \). The participants confirm that the transformation \( \left(\frac{-\ln(0.1)}{\pi}\right)^2 = \frac{x^2}{1-x^2} \) is correct. The final expression for \( x \) is derived as \( x = \sqrt{\frac{1}{\frac{\pi^2}{(\ln(10))^2}+1}} \).
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It is all right so far. What is your problem? You got a simple equation for x2.Captain1024 said:Homework Statement
Solve ##y=\mathrm{exp}(\frac{-x\pi}{\sqrt{1-x^2}})## for x when y = 0.1
Homework Equations
##\mathrm{ln}(e^x)=x##
The Attempt at a Solution
##\mathrm{ln}(0.1)=\frac{-x\pi}{\sqrt{1-x^2}}##
##(\frac{-\mathrm{ln}(0.1)}{\pi})^2=\frac{x^2}{1-x^2}##
Captain1024 said:Homework Statement
Solve ##y=\mathrm{exp}(\frac{-x\pi}{\sqrt{1-x^2}})## for x when y = 0.1
Homework Equations
##\mathrm{ln}(e^x)=x##
The Attempt at a Solution
##\mathrm{ln}(0.1)=\frac{-x\pi}{\sqrt{1-x^2}}##
##(\frac{-\mathrm{ln}(0.1)}{\pi})^2=\frac{x^2}{1-x^2}##
##x=\sqrt{\frac{1}{\frac{\pi ^2}{(\mathrm{ln}(10))^2}+1}}##SammyS said:One thing that will improve the looks of the right hand side. Recognize that −ln(0.1) is ln( (0.1)−1 )
Another thing you might try is to take the reciprocal of both sides of the equation.