Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
06 Jan 2011, 13:37
Kudos
Bookmarks
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 \(m^n^-^1 (m-1)\)
\(m^n - m^n^-^1 = m^n^-^1 (m-1)\)
Mostly I don't understand how \(m^n (1-m^-^1)\) becomes \(m^n^-^1 (m-1)\)
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
06 Jan 2011, 14:14
Kudos
Bookmarks
Cinna
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 \(m^n\): \(m^n(1-m^{-1})=m^n(1-\frac{1}{m})=m^n(\frac{m-1}{m})=\frac{m^n}{m}*(m-1)=m^{n-1}*(m-1)\).
Check Number Theory chapter of Math Book for more on exponents: math-number-theory-88376.html
Hope it's clear.
06 Jan 2011, 15:13
Kudos
Bookmarks
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.
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.
