I've been refreshing my maths over the last couple of weeks, and it's been a challenge since it has been a long time since I was actively using it (20+ years). Anyways, Khan Academy and old textbooks have been a lot of help, but I am stuck on a few things, so I hope you don't mind asking a few basic-ish questions.

I have gone over through the rules of exponents and it seems I got a good grasp on it, as far as challenges go on Khan Academy, but then I opened up an example from the old textbook (has no solution) and I am not sure about it. I need your help here.

This is the task - it says to simplify it:

$$\frac{5^{2n-1} - 25^{n-1}}{125^{n-1}-5^{3n-2}} $$

(original screenshot)

So, what I started doing was making $25^{n-1}$ in the numerator into a ${(5^2)}^{n-1}$ which I got then as $5^{2n-1}$ and from that numerator is 0 and I didn't go through the rest, since numerator == 0 is 0 in result.

But, I have a strong feeling this isn't right and that I made a mistake, but since I have no one to ask and textbook isn't of much help I have to ask you guys for help, guidance here.

Wolfram alpha reports simplified/alternative form as ${(-5)}^{1-n}$ but without further guidance, step-by-step or anything basically.

  • 4
    $\begingroup$ NOTE: $25^{(n-1)}=(5^2)^{(n-1)}=5^{2(n-1)}=5^{2n-2}$. The basic rule being followed is: $$(x^a)^b=x^{ab}$$ $\endgroup$ – Mufasa Mar 22 '15 at 20:41
  • $\begingroup$ How to format mathematics on this web site $\endgroup$ – MJD Mar 22 '15 at 20:42
  • $\begingroup$ Thanks for formatting help! $\endgroup$ – MathAgain Mar 22 '15 at 20:43
  • $\begingroup$ Any time. ${}{}{}$ $\endgroup$ – MJD Mar 22 '15 at 20:59

As noted by Musafa you have: $$ \dfrac{5^{2n-1}-5^{2n-2}}{5^{3n-3}-5^{3n-2}} $$ Now note that $5^{2n-1}=5\times 5^{2n-2}$ and the same for $5^{3n-2}=5 \times 5^{3n-3}$. Using distributivity you can simplifies the fraction and finally you find the result of Wolfram. ( if you have some problem I can help). $$ \dfrac{5^{2n-1}-5^{2n-2}}{5^{3n-3}-5^{3n-2}}=\dfrac{5^{2n-2}(5-1)}{5^{3n-3}(1-5)}= $$ $$ =5^{2n-2-3n+3} (-1)= - (5^{1-n}) $$

  • $\begingroup$ Bear with me, I know this might sound really stupid to you :) But this is giving me a hard time, here is what I did: $\frac{5^{2n-1} - 25^{n-1}}{125^{n-1}-5^{3n-2}} = \frac{5^{2n-1} - (5^2)^{n-1}}{5(5^2)^{n-1}-5^{3n-2}} = \frac{5^{2n-1} - 5^{2n-2}}{25^{2n-2}-5^{3n-2}} = \frac{5^{2n-1} - 5^{2n-21}}{(5^2)^{2n-2}-5^{3n-2}} = \frac{5^{2n-1} - 5^{2n-2}}{5^{4n-4}-5^{3n-2}} = \frac{5^{-1}}{5^{n-2}} $ Not sure still about formatting, so excuse errors! $\endgroup$ – MathAgain Mar 22 '15 at 21:07
  • $\begingroup$ Your last step is wrong. You have the difference of two exponentials and this is NOT the difference of the exponent. I add to my answer.. $\endgroup$ – Emilio Novati Mar 22 '15 at 21:16
  • $\begingroup$ Not only that, but I have also erred in 5^3 at 125 :) Thanks again! $\endgroup$ – MathAgain Mar 22 '15 at 21:31

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