Simplify $\displaystyle \log a^3 - \log a^2$

I'm not sure if this is right but is it either:

a) $\displaystyle = 3\log_{10} a - 2\log_{10} a$

$\displaystyle =\log_{10} a (3-2)$

$\displaystyle =\log_{10} a {1}$

$\displaystyle a=0$

OR

b) $\displaystyle \log a^3 - \log a^2$

$\displaystyle =\log_{10} \frac{a^3}{a^2}$

$\displaystyle =\log_{10} a$

???

To graph $\displaystyle y=e^x$

$\displaystyle -1 \leq x\leq 3$

Is it just the exact same as a normal $\displaystyle y=e^x$ but asymptotic at $\displaystyle x=-1$ and $\displaystyle x=3$?

How can I find the gradient of the tangent at (0,1)?