Solution for $4^{2x+1}-3^{3x+1}=4^{2x+3}-3^{3x+2}$ I'm trying to get a solution for:
$4^{2x+1}-3^{3x+1}=4^{2x+3}-3^{3x+2}$
My main problem is that I don't know how to combine this potencys!
Ive also thought about another function that would bring me same difficulties:
$6^x=36*9.75^{x-2}$
What am I supposed to do?  
 A: $$4^{2x+1}-3^{3x+1}=4^{2x+3}-3^{3x+2}$$
$$4^{2x+1}-4^{2x+3}=3^{3x+1}-3^{3x+2}$$
$$4\cdot4^{2x}-4^34^{2x}=3\cdot3^{3x}-3^23^{3x}$$
$$60\cdot4^{2x}=6\cdot3^{3x}$$
$$10\cdot4^{2x}=3^{3x}$$
$$10\cdot16^{x}=27^{x}$$
$$10=(27/16)^{x}$$
$$\log_{10} 10=\log_{10} (27/16)^{x}$$
$$1=x\log_{10}(27/16)$$
$$x=\frac{1}{\log_{10}27-\log_{10}16}$$
A: For the first one, we have
$$
4^{2x+1}-3^{3x+1}=4^{2x+3}-3^{3x+2} \implies\\
3^{3x+2}-3^{3x+1} = 4^{2x+3}-4^{2x+1} \implies\\
(3-1)3^{3x+1} = (4^2 - 1)4^{2x+1} \implies\\
2\cdot 3^{3x+1} =  5\cdot 3\cdot4^{2x+1} \implies\\
3^{3x} = 5 \cdot 2^{4x+1}
$$
I think that's the simplest we can get it.  From there, I suppose we'd have to solve using logs.
That is, let $\log(x)$ be the logarithm of your choosing.  We have
$$
\log(3^{3x}) = \log(5 \cdot 2^{4x+1}) \implies\\
3x \log(3) = \log(5) + (4x+1)\log(2) \implies\\
(3 \log 3 - 4 \log 2)x = \log 5 + \log 2 \implies\\
x = \frac{\log 5 + \log 2}{3 \log 3 - 4 \log 2} = \frac{\log(10)}{\log(3^3/2^4)} = \frac{1}{\log_{10}27 - \log_{10}16}
$$
