I have this expression $$g = \sqrt {2\ln {{{\sigma ^2}} \over {{{2r} \over {\sqrt a }}x\log x{e^{ - a{{(r + (1/2){\sigma ^2})}^2}\tau /(2{\sigma ^2})}}}}}.$$
I want to put this expression equal to zero so it would be something like
$$0 = {e^{{{{g^2}} \over 2}}} - { \ldots \over \ldots }.$$
${e^{{{{g^2}} \over 2}}}$ I put just to remove squared and $2\ln$ from the right side .
How should I express this correctly?
set the Argument of the logarithm equal to $A$ then we get $$0=\sqrt{2\ln(A)}$$ from here we get $$\ln(A)=0=\ln(1)$$ and we get $$A=1$$
$$g=\sqrt{2\ln\dfrac{\sigma^2}{\dfrac{2r}{\sqrt{a}}x\log xe^{-a(r+\frac{1}{2}\sigma^2)^2\cdot\frac{\tau}{2\sigma^2}}}}\implies e^{\frac{g^2}{2}}=\dfrac{\sigma^2}{\dfrac{2r}{\sqrt{a}}x\log xe^{-a(r+\frac{1}{2}\sigma^2)^2\cdot\frac{\tau}{2\sigma^2}}}$$ $$\implies e^{\frac{g^2}{2}}-\dfrac{\sigma^2}{\dfrac{2r}{\sqrt{a}}x\log xe^{-a(r+\frac{1}{2}\sigma^2)^2\cdot\frac{\tau}{2\sigma^2}}}=0$$ If you wanted, $g=0$ instead, then do that and get $e^{\frac{g^2}{2}}=e^0=1$.
