# How do I solve this fractional indices equation (3^(5x+2))/(9^(1-x))=(27^(4+3x))/(729)

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:
$\frac{{3}^{5x+2}}{{3}^{2\left(1-x\right)}}=\frac{{3}^{3\left(4+3x\right)}}{{3}^{6}}$
Equating the exponents of each numerator:
$5x+2=3\left(4+3x\right)$
$5x+2=12+9x$
$-4x=10$
$x=\frac{10}{-4}=-\frac{5}{2}$
Doing this for the denominator yields a different value of x:
$2\left(1-x\right)=6$
$2-2x=6$
$-2x=4$
$x=-2$
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?
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Nancy Ewing
Recall that the property
${a}^{f\left(x\right)}={a}^{g\left(x\right)}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}f\left(x\right)=g\left(x\right)$
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
${3}^{5x+2-2\left(1-x\right)}={3}^{3\left(4+3x\right)-6}.$
Then you can solve the equation
$5x+2-2\left(1-x\right)=3\left(4+3x\right)-6.$
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 $7\ne 8$ and $5\ne 6$.
###### Did you like this example?
Seamus Mcknight
First subtract the exponents both side and then equate.
${3}^{5x+2-2\left(1-x\right)}={3}^{3\left(4+3x\right)-6}$
$5x+2-2\left(1-x\right)=3\left(4+3x\right)-6\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=-3$