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.

  • $\begingroup$ What have you tried? Are you familiar with exponentiation rules like $a^{m-n}=\frac{a^m}{a^n},(a^n)^m=a^{mn}$ etc? $\endgroup$ – Galc127 Jun 6 '16 at 8:26
  • $\begingroup$ These are two separate equations which are to be solved separatedly, correct? I.e. they will lead to different values of $n$, respectively? $\endgroup$ – Roland Jun 6 '16 at 9:44


$9 =3^2\\25 = 5^2$

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.





$\implies3^{(n-2)-(2-2n)}=3^{2} [a^{(m-n)}=\frac{a^m}{b^m}]$



$\implies n=2$


$=\frac{5^{6-2}}{25^{2-3}} [put(n=2)]$




  • $\begingroup$ It's 6-3 not 6-2.Also it is 25^{-1} not 5^{-1} and also at last its 4+1 not 4-1. $\endgroup$ – Taha Akbari Jun 6 '16 at 8:41

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:

$3^{n-2 -2(1-n)} = 3^2$

  • $\begingroup$ You are right. Just edited. $\endgroup$ – Curious Jun 6 '16 at 8:41

$3^{n-2}=3^{4-2n}$then $n-2=4-2n$ then $n=2$

$5^{3n-3}=5^{2n-3}$then $3n-3=2n-3$ then $n=0$


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