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:
$$\frac{5^{2n-1} - 25^{n-1}}{125^{n-1}-5^{3n-2}} $$
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.
Answer
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).
$$
\dfrac{5^{2n-1}-5^{2n-2}}{5^{3n-3}-5^{3n-2}}=\dfrac{5^{2n-2}(5-1)}{5^{3n-3}(1-5)}=
$$
$$
=5^{2n-2-3n+3} (-1)= - (5^{1-n})
$$
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