What is process/function to cancel base (in value with exponential)? I have this equation (identity?):
$$3^{3x}=3^{2y+1}$$
I understand that this simplifies to:
$$3x=2y+1$$
So, the base (of the exponentials) cancel out. I want to know, what the process/function of this cancelling out is?
I know that you can cancel out exponents by $n$th-rooting the exponents. So by what function do you cancel out the base? 
Please tell me if I should give more information. This is probably a very basic step in dealing with algebraic formulas, but I can't find the specific steps anywhere. I want to know the function so I know what steps to use and how/why they work.
Thanks in advance!
 A: You use the logarithm of the given base. For instance, if you have (in the natural base)
$$e^{f(x)}$$
you can take the natural logarithm of that and get 
$$\ln\left(e^{f(x)}\right)=f(x)\ln(e)=f(x)$$
So in your equation, you take the $\log_3$ on both sides to obtain 
$$\log_3\left(3^{3x}\right)=\log_3\left(3^{2y+1}\right) \leftrightarrow 3x\log_3(3)=(2y+1)\log_3(3) \leftrightarrow$$
$$3x=2y+1$$
A: The function $f(t)=3^t$ is increasing.  So if $f(a)=f(b)$ we have $a=b$.
A: see for basic understanding i can say you that if bases are same and two numbers are in equality then there powers must be equal
$$2^z=2^{33}$$
this means $$z=33$$
but if it is given like $$2^g=15.(2)^{33}$$
then you cannot write $g=33$
for proving this we have to use logarithms. 
A: I feel that you're muddling statements with expressions. Expressions (like $3^{3 \cdot x}$) simplify in a largely mechanical way, whereas statements don't. The statement $3^{3 \cdot x} = 3^{2 \cdot y + 1}$ is equivalent to the statement $3 \cdot x = 2 \cdot y + 1$, and that is proven as follows.
First, we prove that $3^{3 \cdot x} = 3^{2 \cdot y + 1}$ implies that $3 \cdot x = 2 \cdot y + 1$. That comes from the fact that the function $z \mapsto 3^z$ is injective, which requires a lot more machinery to actually prove. From considering the graph of the function, though, it will hopefully be intuitive (hint: it's a strictly increasing function).
Second, we prove that $3 \cdot x = 2 \cdot y + 1$ implies that $3^{3 \cdot x} = 3^{2 \cdot y + 1}$. This is easy, because it follows from the fact that functions map equal values to equal results.
Biïmplication gives the equivalence. It's possible that you only needed one of the two implications (probably the first), but I wanted to make sure.
An alternative to the first part of the proof is to use the $\log_3$ function, as others suggested. We have that $\log_3 3^z = z$ (we could, indeed, have said that $\log_3 3^z$ simplifies to $z$), so applying it to the terms on both sides of the original equation yields two equal terms, specifically $3 \cdot x$ and $2 \cdot y + 1$.
Note that there are plenty of places where there is no direct notion of cancelling out. An example is that the function $x \mapsto x^3-x$ (plot), where it's quite difficult to find anything useful by inverting it.
