How do I solve this fractional indices equation $\frac{3^{5x+2}}{9^{1-x}}=\frac{27^{4+3x}}{729}$? Solve the equation $\frac{3^{5x+2}}{9^{1-x}}=\frac{27^{4+3x}}{729}$
I thought that the best way of approaching this would be to rewrite everything using $3$ as the base of the exponents, hence creating an equivalence which would allow me to equate numerators to numerators, and denominators to denominators. Doing this yields:
$$\frac{3^{5x+2}}{3^{2(1-x)}}=\frac{3^{3(4+3x)}}{3^6}$$
Equating the exponents of each numerator:
$$ 5x+2=3(4+3x) $$
$$ 5x+2=12+9x $$
$$ -4x=10 $$
$$ x = \frac{10}{-4}=-\frac{5}{2} $$
Doing this for the denominator yields a different value of $x$:
$$ 2(1-x)=6 $$
$$ 2-2x=6 $$
$$ -2x=4 $$
$$ x = -2 $$
Why is that I'm obtaining two different values of $x$?
Further to this, the solution in the book states the answer as $x=-3$, what am I doing wrong?

\begin{align}
\frac{3^{5x+2}}{9^{1-x}}=\frac{27^{4+3x}}{729}
\implies & 
\frac{3^{5x+2}}{(3^2)^{(1-x)}}=\frac{(3^3)^{(4+3x)}}{3^6}
\\
\implies & 
\frac{3^{5x+2}}{3^{2\cdot (1-x)}}=\frac{3^{3\cdot(4+3x)}}{3^6}
\\
\implies & 
3^{5x+2-2\cdot (1-x)}=3^{3\cdot(4+3x)-6}
\\
\implies & 
5x+2-2\cdot (1-x)=3\cdot(4+3x)-6
\\
\implies & 
5x+2-2+2x=12+9x-6
\\
&
7x=6+9x
\\
&
-2x=6
\\
&
x=-3
\end{align}
