CVX is not “wrong”. Please read this FAQ so you understand why CVX is doing what it does. I do not know if it is possible for CVX to solve your problem.

The solution to your verbatim problem is not very interesting because it is unattained at x=\infty. But assuming that you actually want to model the inequality

t\geq \log(1+1/y), \ y\geq 0

(which is the same as yours up to linear substitutions), you can write it equivalently as

(y+1)u\geq 1,\ 1-u\geq \exp(-t), \ u,y\geq 0

which is a conic model (one rotated quadratic and one exponential cone). Others will be better at figuring out if that helps to model it in cvx.

@Michal_Adamaszek 's formulation
(y+1)*u >= 1, 1−u >= exp(−t), u,y >= 0
can be handled in CVX with

{1,y+1,u} == rotated_lorentz(1)
1−u >= exp(−t)
y >= 0 % note: 1+y >= 0 will be enforced by rotated_lorentz

@Michal_Adamaszek gets a free virtual beer, plus a free copy of the free version of CVX, for producing a clever comic reformulation into a form which CVX can handle, for at least 3 problems which eluded other forum participants.