Solve the equation $6(3^{2x})=2^{(x+1)}$ To begin this question, I start by using the properties of logarithms:
$$\ln 6 + 2x \ln 3 = (x+1) \ln 2$$
I then distribute the $x+1$ term across ln 2
$$x \ln 2 + \ln2$$
Rewriting the RHS with this, the equation is now:
$$\ln 6 + 2x \ln 3 = x \ln 2 + \ln 2$$
I don't know if this is possible, but in the $2x$ term in the equation, is it possible to put the two back in the exponent of the three and make the three nine? What I mean:
$$2x \ln 3 = x \ln 9$$
For the right hand side, would $x \ln 2 + \ln 2 = x \ln 4$? I believe that is the case but I want to make sure.
I know that the solution is $\frac{-\ln 3}{2 \ln 3 - \ln 2}$, I just do not understand the process of getting to that answer
 A: Another way to approach it for the sake of curiosity:
\begin{align*}
6\cdot3^{2x} = 2^{x+1} & \Longleftrightarrow 3\cdot 9^{x} = 2^{x}\\\\
& \Longleftrightarrow \left(\frac{9}{2}\right)^{x} = \frac{1}{3}\\\\
& \Longleftrightarrow x\ln\left(\frac{9}{2}\right) = -\ln(3)\\\\
& \Longleftrightarrow x = -\frac{\ln(3)}{2\ln(3) - \ln(2)}
\end{align*}
Hopefully this helps!
A: The answer to your first question is yes:  $$2x \ln 3 = x \ln 3^2 = x \ln 9.$$  The answer to your second question is no:  $$x \ln 2 + \ln 2 \ne x \ln 4,$$ because if this were true, it would erroneously imply $$2^{x+1} = 4^x.$$
Instead of taking logarithms so early, it is more useful to simplify first, observing that $6 = 2 times 3$:
$$6(3^{2x}) = 2^{x+1}$$
implies $$2 \cdot 3 \cdot 3^{2x} = 2 \cdot 2^x.$$  Note on the right-hand side, I have used the identity $b^{x + y} = b^x b^y$ with the choice $b = 2$ and $y = 1$.
Then we cancel out the $2$ and absorb the factor of $3$ on the left into the exponent, using the reverse of the same exponent rule we just used earlier:
$$3^{2x + 1} = 2^x.$$
Now we can take logs:
$$(2x+1) \ln 3 = x \ln 2.$$
Rearrange:
$$\frac{2x+1}{x} = \frac{\ln 2}{\ln 3}.$$
Then the left side is
$$2 + \frac{1}{x} = \frac{\ln 2}{\ln 3},$$ hence
$$x = \frac{1}{\frac{\ln 2}{\ln 3} - 2}.$$

Alternatively, we could have obtained this solution from an intermediate point in your calculation, but we would need to observe that
$$\ln 6 = \ln (2 \cdot 3) = \ln 2 + \ln 3.$$
Then substituting this into
$$\ln 6 + 2x \ln 3 = x \ln 2 + \ln 2$$
yields
$$\ln 2 + \ln 3 + 2x \ln 3 = x \ln 2 + \ln 2$$ and canceling then gives
$$(2x + 1) \ln 3 = x \ln 2,$$ which brings us back to the earlier solution above.
A: 
is it possible to put the two back in the exponent of the three and make the three nine? What I mean:$$2x \ln 3 = x \ln 9$$

Yes, this step is correct.

For the right hand side, would $x \ln 2 + \ln 2 = x \ln 4$? I believe that is the case but I want to make sure.

No, this step is wrong, $x\ln 2+\ln2\neq x\ln4$

Rewriting the RHS with this, the equation is now:
$$\ln 6 + 2x \ln 3 = x \ln 2 + \ln 2$$

You can proceed from this step, to solve for $x$. First, move and combine all terms containing $x$
$$\begin{align}(2\ln3-\ln2)x&=\ln2-\ln6\\
\\(2\ln3-\ln2)x&=\ln\left(\frac{2}{6}\right)\\
\\(2\ln3-\ln2)x&=-\ln3\\
\\x&=\frac{-\ln3}{2\ln3-\ln2}\end{align}$$
