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Re: Operations with exponents [#permalink]
DropBear wrote:
Engr2012 wrote:
DropBear wrote:
Hi Everyone,

I have been getting a few problems wrong due to not fully understanding the basics of performing various operations with exponents. Does anyone know of a good blog covering these basics or maybe know of any good problem sets to re-hash this?

Thanks :-D


How about you post a couple of problems that you got wrong and what steps did you falter on? This will give a better understanding of what you are asking.


Hi Engr2012,

it's pretty broad, and just an area I am weak in general I often get the rules mixed up like when do you add exponents, when do you multiply exponents, what happens to the base number when you increase or decrease the exponent. I am sure I could sit there and fumble my way through most problems, but know I need to strengthen that part of my skill set as a whole.

Certainly not representative of everything I am lacking in this area, but an example of something I got wrong recently was this "If 2^2n + 2^2n + 2^2n + 2^2n = 4^24, then n =" I can clearly see this is a simple problem if you follow the steps and required very limited lateral or outside the box thinking but I sat there struggling with the steps and got it wrong.

Just through someone along the line might have made a review or work set covering all the major operations with exponents.



Check here:

strategy-for-exponents-and-roots-138017.html#p1116730
strategy-for-exponents-and-roots-138017.html#p1116916
new-tough-and-tricky-exponents-and-roots-questions-125956.html

Just to summarize a few concepts for exponents:

\(A^m * A^n = A^{m+n}\)

\(\frac{A^m}{A^n} = A^{m-n}\) ---> this can be visualized as follows:

\(\frac{1}{A^n}\) = \(A^{-n}\)

\(A^m*A^{-n} = A^{m-n}\)

\((A^m)^n\) = \(A^{mn}\)

\(\sqrt{A} = A^{1/2}\)

\(\sqrt[3]{A} = A^{1/3}\)

The biggest and the most important step is to break down the 'bigger' numbers into their corresponding prime factors. This way you will be able to reach the correct answer.

As for the question you posted, \(2^{2n}+2^{2n}+2^{2n}+2^{2n}\) = \(4*2^{2n}\) = \(2^2 * 2^{2n}\)= \(2^{2+2n}\) = \(4^{24}\)

\(2^{2+2n}\) = \(4^{24}\) ---> \(2^{2+2n}\) = \((2^2)^{24}\) ---> \(2^{2+2n}\) = \(2^{2*24}\) ---> equate the powers of same "bases", we get

2+2n=48 ---> n =23

Hope this helps.



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