Intuition: Why will $3^x$ always eventually overtake $2^{x+a}$ no matter how large $a$ is? I have a few ways to justifiy this to myself.
I just think that since $3^x$ "grows faster" than $2^{x+a}$, it will always overtake it eventually. Another way to say this is that the slope of the tangent of $3^x$ will always eventually be greater for some $x$ than that of $2^{x+a}$, so that the rate of growth at that $x$ value will be greater for $3^x$, so at that point it's only a matter of "time" before it overtakes $2^{x+a}$.
Another way I think about it is that the larger $x$ becomes, the closer the ratio $x:(x+a)$ comes to $1$, in other words $\lim_{x \to \infty} (\frac{x}{x+a}) = 1$, so that the base of the exponents is what really matters asymptotically speaking.
However, I'm still not completely convinced and would like to rigorize my intuition. Any other way of thinking about this would be very helpful, whether geometric (visual intuition is typically best for me), algebraic, calculusic...anything.
This came to me because I was trying to convince myself that $\frac{b \cdot 2^x}{3^x}$ goes to $0$ no matter how large $b$ is, and I realized that $b$ can be thought of as $2^a$, and that it might be easier to see this way, but if you have some intuition fo this form: $\frac{b \cdot 2^x}{3^x}$, I welcome it also.
 A: Consider
$\dfrac{2^a}{3^a}= \left(\dfrac{2}{3}\right)^a$
Since $\frac{2}{3}$ is less than $1$, the larger $a$ gets, the smaller $\left(\frac{2}{3}\right)^a$ gets (closer to $0$).
A: $2^{x+a} = 2^x 2^a$, so the problem reduces down to $3^x$ surpassing $c 2^x$ for any positive $c$. Dividing both sides by $2^x$ results in $(3/2)^x$ surpassing $c$, which can be achieved using logarithm and monotonicity.
A: Indeed for all constants $C$ we have that there is an $N$ such that for all $x>N$, $3^x>C\cdot 2^x$. This is often abbreviated $2^x=O(3^x)$. This answers your question as a special case because $2^{x+a}=2^x\cdot 2^a$.
To see that it is true, take logarithms to get
$$x\ln 3>x\ln 2+\ln C$$
or
$$x(\ln 3-\ln 2)>C$$
so
$$x>C/(\ln 3-\ln 2)$$
We can take $N$ to be any number larger than the right hand side to get the desired result.
A: $3^x<2^{x+a}\iff$
$3^x<2^x\cdot2^a\iff$
$\frac{3^x}{2^x}<2^a\iff$
$1.5^x<2^a\iff$
$x<\log_{1.5}2^a\iff$
$x<a\log_{1.5}2\iff$
$x<a\log_{1.5}2\implies$
$x<1.8a$
So no matter how large $a$ is, when $x$ is larger than $1.8a$, $3^x$ is larger than $2^x$.
