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Would someone explain how or perhaps where I could learn how

\(m^n - m^n^-^1 = m^n^-^1 (m-1)\)

Mostly I don't understand how \(m^n (1-m^-^1)\) becomes \(m^n^-^1 (m-1)\)

Negative powers: \(a^{-n}=\frac{1}{a^n}\)

Operations involving the same bases: Keep the base, add or subtract the exponent (add for multiplication, subtract for division) \(a^n*a^m=a^{n+m}\)

\(\frac{a^n}{a^m}=a^{n-m}\)

Now, as for your problem: you can directly factor out \(m^{n-1}\) from \(m^n-m^{n-1}\) and get \(m^{n-1}*(m-1)\) as \(m^{n-1}*(m-1)=m^{n-1}*m-m^{n-1}=m^n-m^{n-1}\).

Or the long way: \(m^n-m^{n-1}=m^n-m^n*m^{-1}\) -> factor out \(m^n\): \(m^n(1-m^{-1})=m^n(1-\frac{1}{m})=m^n(\frac{m-1}{m})=\frac{m^n}{m}*(m-1)=m^{n-1}*(m-1)\).

I certainly see the short method and understand that, I'm still stuck though on the long hand.

When I get to this step, \(m^n (\frac{m-1}{m})\) I don't recognize the rules that let us move the division to the other part of the forumla to finish it.

I certainly see the short method and understand that, I'm still stuck though on the long hand.

When I get to this step, \(m^n (\frac{m-1}{m})\) I don't recognize the rules that let us move the division to the other part of the forumla to finish it.

You can manipulate in many ways: \(\frac{xy}{z}=x*\frac{y}{z}=y*\frac{x}{z}=xy*\frac{1}{z}=xyz^{-1}=...\). So, \(m^n (\frac{m-1}{m})= \frac{m^n(m-1)}{m}=\frac{m^n}{m}*(m-1)=m^{n-1}*(m-1)\).

I'm using the MGMAT Number Properties book and I'm hoping someone can please explain a concept for me.

Going to and from Distributed Form and Factored Form...

a^b-a^(b-1) turns into a^b(1-a^(-1)) = a^(b-1)(a-1) I understand the first step, but how do you get from a^b(1-a^(-1)) to a^(b-1)(a-1)?

Sorry for not using the math tag. The equation in the exponent seems to mess up the formatting. Thanks in advance!

The only difference is the names of variables:

You can directly factor out \(a^{b-1}\) from \(a^b-a^{b-1}\) and get \(a^{b-1}*(a-1)\) as \(a^{b-1}*(a-1)=a^{b-1}*a-a^{b-1}=a^b-a^{b-1}\).

Or the long way: \(a^b-a^{b-1}=a^b-a^b*a^{-1}\) -> factor out \(a^b\): \(a^b(1-a^{-1})=a^b(1-\frac{1}{a})=a^b(\frac{a-1}{a})=\frac{a^b}{a}*(a-1)=a^{b-1}*(a-1)\).
_________________

I'm using the MGMAT Number Properties book and I'm hoping someone can please explain a concept for me.

Going to and from Distributed Form and Factored Form...

a^b-a^(b-1) turns into a^b(1-a^(-1)) = a^(b-1)(a-1) I understand the first step, but how do you get from a^b(1-a^(-1)) to a^(b-1)(a-1)?

Sorry for not using the math tag. The equation in the exponent seems to mess up the formatting. Thanks in advance!

The only difference is the names of variables:

You can directly factor out \(a^{b-1}\) from \(a^b-a^{b-1}\) and get \(a^{b-1}*(a-1)\) as \(a^{b-1}*(a-1)=a^{b-1}*a-a^{b-1}=a^b-a^{b-1}\).

Or the long way: \(a^b-a^{b-1}=a^b-a^b*a^{-1}\) -> factor out \(a^b\): \(a^b(1-a^{-1})=a^b(1-\frac{1}{a})=a^b(\frac{a-1}{a})=\frac{a^b}{a}*(a-1)=a^{b-1}*(a-1)\).

Thanks for the response! Man, that long way really is long. Are there any times when I will actually have to do the long way, or can I just sort of skip all that and memorize the format of the factored form? I guess what I'm really asking is, are there any conditions that I should be aware of for when I can or can't use the shortcut?

Thanks for the response! Man, that long way really is long. Are there any times when I will actually have to do the long way, or can I just sort of skip all that and memorize the format of the factored form? I guess what I'm really asking is, are there any conditions that I should be aware of for when I can or can't use the shortcut?

Second approach looks long just because every step is written. You can use first approach if you like it more. Anyway, I'd advice to study and understand fundamentals rather than to memorize some specific "formats".
_________________

Second approach looks long just because every step is written. You can use first approach if you like it more. Anyway, I'd advice to study and understand fundamentals rather than to memorize some specific "formats".

I hear you on that. The long formula isn't as bad as I thought it would be. I tried playing around with it, going forward and backwards, and subbing in numbers, and it seems straight-forward enough. But the real test will be when I need to recognize and apply it in a problem (still haven't run into any problems that require it). Thanks again!

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