given the functions $\displaystyle g(x)=e^x-2, x=(a,infinity)$
$\displaystyle
h(x)=ln(lnx), x>1
$
Find the smallest value of $\displaystyle a$ in exact form such that the composite function $\displaystyle hg$ exists.
Assuming that you are trying to find $\displaystyle \displaystyle h\left(g(x)\right)$...
$\displaystyle \displaystyle h\left(g(x)\right) = \ln{\left[\ln{\left(e^x - 2\right)}\right]}$.
A logarithm is only defined for positive values of the independent variable, so the innermost logarithm can only be evaluated where $\displaystyle \displaystyle e^x - 2 > 0$.
But since another logarithm will be taken, you can only accept those values for which the innermost logarithm is positive. So where $\displaystyle \displaystyle e^x - 2 > 1$.
Solve for $\displaystyle \displaystyle x$.