How to simplify $\ln |\ln y | = \ln|x|+C$?

Question

Let's say after solving an ODE, the solution I ended up with is as follows,

$$\ln |\ln y | = \ln | x | + C$$

and I want to simplify it in a manner where I can get $y$ by itself. How would I go about that?

Answer 1

Exponentiate once and absorb the sign from removing the absolute values and the exponential of the constant into a new constant, $C_1=sign(x_0)sign(\ln y_0)e^C$, then $$ \ln y=C_1x $$ which then obviously implies $$ y=e^{C_1x}. $$

Answer 2

From $\ln |\ln y | = \ln | x | + C$, we get $|\ln y|=e^C|x|$ from which $y=e^{\pm e^Cx}$. Since $e^{e^C}>1$ and $0<e^{-e^C}<1$, the general solution is $y=a^{x}$ with $a>0$ and $a\neq 1$.