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?
 A: Recall that the property
$$a^{f(x)} = a^{g(x)} \implies f(x) = g(x)$$
holds because the exponential function is injective. Therefore, you always need to rewrite both sides of the equation as powers of $a$ before you can equate the exponents.
In your situation, you can apply the property:
$$\frac{a^m}{a^n} = a^{m - n}$$
and thus write
$$3^{5x+2 - 2(1 - x)} = 3^{3(4 + 3x) - 6}.$$
Then you can solve the equation
$$5x+2-2(1-x) = 3(4+3x)-6.$$
Now, if you're still wondering why you can't just equate the exponents of the numerators and those of the denominators, let me show you a simple counterexample to explain why that might not work in general:
$$\frac {3^7} {3^5} = \frac {3^8} {3^6}$$
holds because if you compute the quotients they are both equal to $9$, but $7 \neq 8$ and $5 \neq 6$.
A: First subtract the exponents both side and then equate.
$$3^{5x+2-2(1-x)}=3^{3(4+3x)-6}$$
$$5x+2-2(1-x)=3(4+3x)-6 \implies x=-3$$
A: \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}
A: You cannot just equate the numerator and denominator here, because they are not coprime.
Use that $$\frac{3^{5x+2}}{3^{2(1-x)}}=3^{5x+2-2(1-x)}$$ and $$\frac{3^{3(4+3x)}}{3^6}=3^{3(4+3x)-6}$$
then solve
$$5x+2-2(1-x)=3(4+3x)-6$$
A: HINT
$$\dfrac{a^m}{a^n}=a^{m-n}$$
You need to subtract exponents. Notice that the bases are the same.
