If $3^{33}+3^{33}+3^{33}=3^{x}$. Solve for $x$. If $3^{33}+3^{33}+3^{33}=3^{x}$. Solve for $x$.

So we have:

$$3^{33}+3^{33}+3^{33}=3^{x}$$

I added the left side and obtained:

$3(3^{33})=3^{x}$

The problem I have is that extra $3$. If not, I could have said $x=33$. Any hints in how to proceed with this problem?
$ 3(3^{33})=  3^{x} $, then $ x = 34 $ .
$3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3*3=3^x$