Exponential growth and decay for the decomposition of nitrogen(V) oxide 
Experiments show that if the chemical reaction
$$\mathrm{N_2O_5 \to 2\,NO_2 + \frac{1}{2}\,O_2}$$
takes place at $45\,\rm^\circ C$, the rate of the reaction of dinitrogen pentoxide is proportional to its concentration as follows:
$$-\frac{d[\mathrm{N_2O_5}]}{dt} = 0.0005[\mathrm{N_2O_5}]$$
a) Find an expression for the concentration $[\mathrm{N_2O_5}]$ after $t$ seconds if the initial concentration is $C$.
b) How long will the reaction take to reduce the concentration of $\mathrm{N_2O_5}$ to $90\%$ of its original value?

Part a
I simply used the instantaneous rate of $0.0005$ as the constant $k$ (referring to the rate of reaction) and, since $C$ is a constant, I utilized it as the initial value in the formula:
$$y(t)=y(o)e^{kt}=Ce^{kt}$$
I replaced $k$ for $0.0005$: the book has a negative sign in front of this value, I am curious as to why since the equal sign would indicate otherwise, no? Nonetheless, the equation:
$$Ce^{-0.0005(t)}$$
Please comment on my reasoning.
Part b
I simply utilized the above equation and set it equal to the decimal form of 90% with a negative sign since the question  said reduced
$$Ce^{-0.0005(t)} = -0.9$$
However, I really don't understand conceptually why $90\%$ can be considered $0.9$ of the concentration. I just need to clear this up in my mind.
From that point I am pretty lost why the $C$ appears on both sides of the equation and why it disappears.
Answer from the book:

$$y(t)=Ce^{-0.0005}=0.9C\rightarrow Ce^{-0.0005}=0.9C\rightarrow e^{-0.0005}=0.9$$

What happened to $C$?
 A: First of all, we denote the concentration of $\mathrm{N_2O_5}$ at time $t$ with $y(t)$. Then your governing equation is 
$$ - {{dy(t)} \over {dt}} = ky(t)$$
The solution to this simple ordinary differential equation (ODE) is
$$y(t) = A{e^{ - kt}}$$
where $A$ is an arbitrary constant. You can simply check this by substituting into the ODE. Your first mistake was to take $y(t) = A{e^{kt}}$ as the solution of the ODE, neglecting the minus sign. Now, we have the initial condition
$$y(0) = C$$
where $C$ is a known constant which is the initial concentration of $\mathrm{N_2O_5}$. Then our solution becomes
$$y(t) = C{e^{-kt}}$$
So the solution of Part a is finished since we have found the concentration of $\mathrm{N_2O_5}$ at time $t$ in terms of its initial concentration. Part b wants to know at what time the concentration is $90\%$ of the initial concentration, i.e.
$$y({t^*}) = {{90} \over {100}}y(0) = {{90} \over {100}}C = 0.9C$$
where ${t^*}$ is the time we want to find. So we may write
$$C{e^{ - k{t^*}}} = 0.9C$$
Divide the above equation by $C$
$${e^{ - k{t^*}}} = 0.9$$
taking natural logarithm from both sides we have
$$ - k{t^*} = \ln (0.9)$$
and finally, solving for ${t^*}$ leads to
$${t^*} =  - {{\ln (0.9)} \over k}$$
