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I have had some issues with the following two equations:
$$ \frac{3^{n-2}}{9^{1-n}}=9$$ $$\frac{5^{3n-3}}{25^{n-3}}=125$$
If anyone could work them out step by step that would be awesome. I keep arriving at the wrong answer and believe it could be related to the addition of exponents towards the end of the question.
note:$n$ is different in each equation.
Hint:
$9 =3^2\\25 = 5^2$
Full solution:
$$\begin{align}\frac{3^{n-2}}{9^{1-n}} = 9\\\frac{3^{n-2}}{3^{2-2n}} = 3^2 && \text{(Using $9 = 3^2$ and $(a^b)^c = a^{bc}$)}\\3^{(n-2) - (2-2n)} = 3^2 && \text{($\frac{a^b}{a^c} = a^{b-c}$)}\\3n-4 = 2 && \text{(Equating exponents)}\\3n = 6\\n=2\end{align}$$
The second question can be done the same way.
$\frac{3^{n-2}}{9^{1-n}}=9$
$\implies\frac{3^{n-2}}{3^{2(1-n)}}=9$
$\implies\frac{3^{n-2}}{3^{2-2n}}=9$
$\implies3^{(n-2)-(2-2n)}=3^{2} [a^{(m-n)}=\frac{a^m}{b^m}]$
$\implies3^{3n-4}=3^2$
$\implies{3n-4}=2$
$\implies n=2$
$\frac{5^{3n-3}}{25^{n-3}}$
$=\frac{5^{6-2}}{25^{2-3}} [put(n=2)]$
$=\frac{5^4}{5^{-1}}$
$=5^{4-1}$
$=5^3=125$
Recall that the first equation is equal to:
$\dfrac{3^{n-2}}{9^{1-n}}=9 \implies \dfrac{3^{n-2}}{3^{2(1-n)}}=3^2$
This is equal to:
$3^{n-2 -2(1-n)} = 3^2$
$3^{n-2}=3^{4-2n}$then $n-2=4-2n$ then $n=2$
$5^{3n-3}=5^{2n-3}$then $3n-3=2n-3$ then $n=0$
