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factoring exponents

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by Night reader » Sat Jan 29, 2011 3:02 pm
SFtraveler wrote:Please show each step in factoring: 5^k-5^(k-1)
5^k -5^(k-1) = 5^k(1 - 1*(-1/5)) = 6*5^k/5
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by Everest » Sat Jan 29, 2011 3:21 pm
5^k-5^(k-1) => 5^K - 5^k * 5^ -1 = 5^k (1 - 1/5) = 5^K * 4/5 = 4 (5^(k-1))
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by Everest » Sat Jan 29, 2011 3:43 pm
Night reader wrote:
SFtraveler wrote:Please show each step in factoring: 5^k-5^(k-1)
5^k -5^(k-1) = 5^k(1 - 1*(-1/5)) = 6*5^k/5

Not sure how 5^k -5^(k-1) becomes 5^k(1 - 1*(-1/5))

It should be 5 ^k - 5^k/5 = 5^k(1-1/5) = 4/5 * 5^k or 4 * 5^k-1
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by SFtraveler » Sat Jan 29, 2011 3:49 pm
I show the answer being (4/5) 5^k.

Everest, thank you for your answer. I messed up at 5^(k-1) = 5^k * 5^-1. I found the exponent rule (x^a)/(x^b) = x^(a-b) that covers doing that step.

Night reader, your answer is incorrect.
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by Night reader » Sat Jan 29, 2011 4:16 pm
SFtraveler wrote:I show the answer being (4/5) 5^k.

Everest, thank you for your answer. I messed up at 5^(k-1) = 5^k * 5^-1. I found the exponent rule (x^a)/(x^b) = x^(a-b) that covers doing that step.

Night reader, your answer is incorrect.
@SFTraveler I found this problem as well as the solution defined by you as correct one not confident...

-a^x is the same as (-a)^x, but not the same as -(a)^x

-5^(k-1) = -5^(k+ (-1)) = -5^k * -5^(-1) = -5^k * -1/5 = (5^k)/5

and 5^k -5^(k-1) = 6/5 *5^k
----------------------------

5^k -(5^(k-1)) = 4/5 *5^k

what's the source of this question?

check point
k=3
5^3 - 5^(3-1) = 125 -5^2 = 125+25= 150 <> 150= 6/5 * 5^3 OR 6*125/5 = 6*25 and not 4/5*5^3 :)

5^3 -(5^(3-1))= 125 -(5^2) = 125-25 = 100 <> 4/5 * 5^3

check the statement in your book again, and tell us the source
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by SFtraveler » Sat Jan 29, 2011 4:39 pm
Source is beatthegmat.com flashcards (factoring exponents example)
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by Night reader » Sat Jan 29, 2011 4:46 pm
SFtraveler wrote:Source is beatthegmat.com flashcards (factoring exponents example)
Thank you, I am attaching the original slide from the BTG flashcards.

and I think we should ask Eric to have the above discussed statement corrected symbol-wise :)
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BTG flashcards -exponents.JPG
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by beatthegmat » Sat Jan 29, 2011 6:04 pm
Thank you for pointing this out! I'll try to get this fixed shortly.
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