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Subtracting Exponents with Same Base but Different Exponent 
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#1
Nov1905, 08:58 PM

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How do you do this problem (without a calculator) that looks deceivingly simple, yet is utterly confusing. (2 ^ 30  2 ^ 29)/2 = ?. The answer is
2 ^ 28, but how was this done without using a calculator. The furthest I can take this without using a calculator is 2 ^ 29  2 ^ 28 


#2
Nov1905, 09:05 PM

P: 461

2 ^ 29  2 ^ 28 = 2^28(21) = 2^28(1) = 2^28



#3
Nov2005, 08:23 PM

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2^{30} 2^{29}= 2(2^{29}) 2^{29}=
(21)2^{29}= 2^{29} so (2^{30}2^{29})/2= 2^{29}/2= 2^{28}. 


#4
Jul3006, 02:22 PM

P: 3

Subtracting Exponents with Same Base but Different Exponent
I am trying to work this actual problem that is in my GRE practice book. Looking at the way it is worked i still need help. Please let me know if you could help in any way.



#5
Jul3006, 03:03 PM

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Complete solutions have been posted above. What specific help are you looking for? If you explain which concept you're having difficulty with, someone might be able to help you better.
Edit : If your problem is entirely unrelated to the prior content of this thread, state the COMPLETE question in a new thread. 


#6
Jul3006, 03:17 PM

P: 3

Well i understand how the orginal problem 2^302^29/2 becomes 2^29 2^28 because you subract 2^1 from each exponent. Why isn't the 2 at the bottom cancelled out? And where is the (21) coming from? The rule in the GRE manual for example 5^7/5^4 becomes 5^3 thus cancelling the denominator of 5. Please help me clear up any confusions.



#7
Jul3006, 05:39 PM

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Let's try being more explicit then:
(2^{30}2^{29})/2 Step one: cancel the two in the denominator. TO do this, you divide each term by 2, meaning (for this specific scenario), you subtract one from each exponent 2^{29}2^{28} Step two: Factor 2^{28} from all the terms. 2^{28}(2^{1}2^{0}) Step three: Remembering some basic exponent rules, we clean up the equation a bit (specifically a number to the power of 1 or 0) 2^{28}(21) Step four: Finish solving by calculating 21 2^{28}(1) = 2^{28} 


#8
Apr511, 07:58 PM

P: 1

i have the same problem with the problem above. this is the problem:
3^(n+2) / [3^(n+3)  3^(n+1)] the answer key says that the answer is 1/3 but when I solved for it, the answer is 3/8.. what is really the correct answer and how will it be solved? 


#9
Apr611, 06:17 AM

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If you only have powers of 3, you cannot possibly have a factor of [itex]8= 2^3[/itex]!
The correct answer is 1/3 as you were told write [itex]3^{n+2}[/itex] as [itex]3(3^{n+1})[/itex] and cancel. 


#10
Apr611, 12:48 PM

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[tex]\frac{3^{n+2}}{3^{n+3}3^{n+1}} = \frac{3 3^{n+1}}{3^2 3^{n+1}3^{n+1}} = \frac{3}{3^23^0} = \frac{3}{91} = \frac{3}{8}[/tex] Perhaps HallsofIvy and the person who wrote the book's answer key missed the "3^(n+1)" in the denominator? 


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