# Exponential Equations with Fractions

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.

• 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? – Galc127 Jun 6 '16 at 8:26
• These are two separate equations which are to be solved separatedly, correct? I.e. they will lead to different values of $n$, respectively? – Roland Jun 6 '16 at 9:44

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.

$\frac{3^{n-2}}{9^{1-n}}=9$

$\implies\frac{3^{n-2}}{3^{2(1-n)}}=9$

$\implies\frac{3^{n-2}}{3^{2-2n}}=9$

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

$\implies3^{3n-4}=3^2$

$\implies{3n-4}=2$

$\implies n=2$

$\frac{5^{3n-3}}{25^{n-3}}$

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

$=\frac{5^4}{5^{-1}}$

$=5^{4-1}$

$=5^3=125$

• 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. – 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:

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

• You are right. Just edited. – 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 $n=0$