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?
Best Answer
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 $5 \neq 6$.