Please Explain $\lg(T(N)) = 3 \lg N + \lg a$ is equivalent to $ T(N) = aN^3$ I'm reading Algorithms by Kevin Wayne and Robert Sedgewick. 
They state that:
$\lg(T(N)) = 3 \lg N + \lg a $
(where $a$ is constant) is equivalent to 
$T(N) = aN^3$
I know that $\lg$ means a base $10$ logarithm and that $\lg(T(N))$ means the index of the power to which $10$ must be raised to produce $T(N)$ but I'd like some help understanding how to get from the first equation to the second. 

Simply raise $10$ to the power of both sides of the equation:
$\large{10^{\log {T(N)}}=10^{3\log N +\log a}=10^{3\log N}\cdot10^{\log a}=(10^{log N})^3\cdot10^{\log a}}$
Since by definition $\log b = c \iff 10^c=b$, it follows that $10^{\log b}=b$, and thus
$T(N)=N^3 \cdot a$.
