Given that X follows Uniform $[0,1]$ find the PDF of $Y=\text{log } X$
Calculate the distribution of $$Y:F(y) = P(Y<y)$$ In this case, we have $$F(y)=P(\log X<y)=P(X<e^y) =\int_0^{e^y}dt$$
Completing the calculating gives $$PDF=f(y)=e^y, 0<y<\infty$$
And this does not have area under the curve as $1$? In, fact, it will be infinite. And when I do a transformation shouldn't the PDF already be normalized?
How do I resolve this?
Adding on to the other answer, the flaw in your reasoning lies at: "$\displaystyle P(X < e^y) = \int_0^{e^y}dt$". This is false, as the PDF of $X$ is not $\mathbf1_{\mathbb R_+}$, but $\mathbf 1_{[0,1]}$. The correct continuation would be, for $y\in \mathbb R$: $$ F(y) = P(\log X < y) = P(X < e^y) = \int_{-\infty}^{e^y}\mathbf 1_{[0,1]}dt = \int_0^{\min(e^y,1)}dt = \min(e^y,1).$$ Thus the PDF of $Y$ would be: $$g(y) = e^y, \, -\infty < y < 0.$$
Because $X<1$ almost surely, we have $Y<\log(1)=0$ almost surely. Thus, for positive $y$, we have $$ P(Y<y)=P(X<e^y)\ge P(X<1)=1 $$ The formula $P(Y<y)=e^y$ is only true when $y$ is not positive.
