Exponential distribution - With Half life

$$X \sim Exp(\lambda)$$ and it has a half life of $$1$$ (year) (lets say you have a set of x's and they decay and after 1 year you have half of the x's you had)

Questions:

1. Calculate the time when the expectancy $$\mathbb{E}$$ of the number of X's left is 10% of the number of X's at the beginning.

2. If you have 1024 X's at the beginning, calculate the time when the expectancy of the number of X's is $$1$$

3. The probability that no x from all the 1024 x's were gone after the time you calculate at #2

My Try:

1. if the half life is $$1$$ then: $$\frac{\ln{2}}{\lambda} = 1$$ so: $$\lambda = \ln{2}$$
$$X \sim \text{Exp}(\ln{2})$$ $$f(x) = \ln(2) \cdot e^{-\ln(2)x} = \frac{\ln(2)}{2^x}$$
Now I need to find what time it takes for X to be 10% so $$0.1$$ of the original. the problem is that I don't know how much X's were in the first place. maybe 1 is the total percentage? (100%) ?

so: $$\frac{1}{\frac{ln2}{2^x}} = 0.1 \Leftrightarrow 2^x = \ln(2) \cdot 0.1 \Leftrightarrow x = \log_{2}(\ln(2) \cdot 0.1) =$$ ? I get negative result.. where am I wrong here?

1. $$1024 \cdot \ln(2) \cdot e^{-x} = 1$$ and so the answer is $$x = \ln(\ln(2) \cdot 1024) = 6.565$$

2. If we calculated the time where the expected number of x's is 1 then after infinite many years it would decay to completely $$0$$ no? I don't get this question...

Thank you very much!

An "exponentially distributed lifetime" means that $$\mathsf E_t(X)=X_0\mathrm e^{-\lambda t}$$ .. the expected amount of $$X$$ at time $$t$$ equals the original amount ($$X_0$$) times the decay coefficient.   We usually simply write $$X(t)$$ or $$X_t$$ for this expected amount.
The half-life, $$t_{1/2}$$, of this decay process is the time the expected amount equals half the original amount. So, $$t_{1/2}=1\text{yr}$$ indeed means, that $$\tfrac 12=\mathrm e^{-\lambda (1\text{yr})}$$, or $$\lambda = \ln (2)\, \text{yr}^{-1}$$ .
Similarly, you need to find $$t$$ such that $$0.10=\mathrm e^{-\lambda t}$$, or \begin{align}t ~&=~-\ln(0.10)/\ln(2)\\&\approx~3.32\end{align}