I know the answer is:
$$f_T(t) = \frac{d}{dt} F_{\ln t}(\ln(t)) = f_{\ln t}(\ln t) \frac{1}{t}$$
which you can get by using cdfs and taking the derivative.
$$F_T(t) = P(T \le t) = P(\ln(T) \le \ln(t)) = F_{\ln t}(\ln(t))$$
My question: I am having difficulty getting the answer using this alternative approach which uses the formula:
$$f_Y(y) = f_X(x) \frac{dx}{dy} \quad \text{with} \quad x = g^{-1}(y)$$
My attempt:
Here $Y = T$ and $X = \ln T$ so we have
$$f_T(t) = f_{\ln T}(x) \frac{dx}{dt} $$
and now I have to find $x = g^{-1}(t)$ but at this point I'm stuck. I know $x = lnt$ but I'm having issues writing this out formally.
It's tripping me out because we go from $lnt$ to $t$ which is a simpler function. Thanks for your help and patience.