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easy exponents

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by deagez » Sun Jul 12, 2009 10:53 am
what is a quick way to think about the following?

2^5+2^5+3^5+3^5+3^5
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by truplayer256 » Sun Jul 12, 2009 11:00 am
2^5+2^5+3^5+3^5+3^5

2*(2^5)+3*(3^5)=2^(5+1)+3^(5+1)=2^6+3^6
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by mehravikas » Wed Jul 15, 2009 10:30 pm
Shouldn't it be - 2^6+3^7
truplayer256 wrote:2^5+2^5+3^5+3^5+3^5

2*(2^5)+3*(3^5)=2^(5+1)+3^(5+1)=2^6+3^6
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by tohellandback » Wed Jul 15, 2009 10:36 pm
mehravikas wrote:Shouldn't it be - 2^6+3^7
truplayer256 wrote:2^5+2^5+3^5+3^5+3^5

2*(2^5)+3*(3^5)=2^(5+1)+3^(5+1)=2^6+3^6
no. coz 3*3^5=3^(5+1)=3^6
The powers of two are bloody impolite!!
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Meharvikas......

by struggling_guy2001 » Wed Jul 15, 2009 10:49 pm
Hi,


Wanna to remind you the very basic formula or property behind this...

(a) ^ (x) * (a) ^ (y) = (a) ^ (x+y).
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by mehravikas » Thu Jul 16, 2009 11:59 am
In the question 3^5 appears thrice: 3^5+3^5+3^5
tohellandback wrote:
mehravikas wrote:Shouldn't it be - 2^6+3^7
truplayer256 wrote:2^5+2^5+3^5+3^5+3^5

2*(2^5)+3*(3^5)=2^(5+1)+3^(5+1)=2^6+3^6
no. coz 3*3^5=3^(5+1)=3^6
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by tohellandback » Thu Jul 16, 2009 12:02 pm
mehravikas wrote:In the question 3^5 appears thrice: 3^5+3^5+3^5
tohellandback wrote:
mehravikas wrote:Shouldn't it be - 2^6+3^7
truplayer256 wrote:2^5+2^5+3^5+3^5+3^5

2*(2^5)+3*(3^5)=2^(5+1)+3^(5+1)=2^6+3^6
no. coz 3*3^5=3^(5+1)=3^6
yes and thats why it is 3 * 3^5. Now 3* 3^5=3^6
The powers of two are bloody impolite!!
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by mehravikas » Thu Jul 16, 2009 12:08 pm
Are you trying to say the occurrence of an integer does not matter?

2^5+2^5 = 2^6

3^5+3^5+3^5 = 3^6

tohellandback wrote:
mehravikas wrote:In the question 3^5 appears thrice: 3^5+3^5+3^5
tohellandback wrote:
mehravikas wrote:Shouldn't it be - 2^6+3^7
truplayer256 wrote:2^5+2^5+3^5+3^5+3^5

2*(2^5)+3*(3^5)=2^(5+1)+3^(5+1)=2^6+3^6
no. coz 3*3^5=3^(5+1)=3^6
yes and thats why it is 3 * 3^5. Now 3* 3^5=3^6
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by tohellandback » Thu Jul 16, 2009 12:13 pm
mehravikas wrote:Are you trying to say the occurrence of an integer does not matter?

2^5+2^5 = 2^6

3^5+3^5+3^5 = 3^6

tohellandback wrote:
mehravikas wrote:In the question 3^5 appears thrice: 3^5+3^5+3^5
tohellandback wrote:
mehravikas wrote:Shouldn't it be - 2^6+3^7
truplayer256 wrote:2^5+2^5+3^5+3^5+3^5

2*(2^5)+3*(3^5)=2^(5+1)+3^(5+1)=2^6+3^6
no. coz 3*3^5=3^(5+1)=3^6
yes and thats why it is 3 * 3^5. Now 3* 3^5=3^6
Ummm I am trying to say that its a formula for exponents which can be explained in multiple ways. And I suggest you take any book even the OG and go through the exponents basics once. Should not take more than 20 minutes.
The powers of two are bloody impolite!!
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by mehravikas » Thu Jul 16, 2009 12:23 pm
Thanks..Basically I got confused in a step that was not explained in the answer explanations:

3^5+3^5+3^5 = 3^5(1 + 1 + 1) = 3^5 * 3^1 = 3^6

Cheers,
Vikas
tohellandback wrote:
mehravikas wrote:Are you trying to say the occurrence of an integer does not matter?

2^5+2^5 = 2^6

3^5+3^5+3^5 = 3^6

tohellandback wrote:
mehravikas wrote:In the question 3^5 appears thrice: 3^5+3^5+3^5
tohellandback wrote:
mehravikas wrote:Shouldn't it be - 2^6+3^7
truplayer256 wrote:2^5+2^5+3^5+3^5+3^5

2*(2^5)+3*(3^5)=2^(5+1)+3^(5+1)=2^6+3^6
no. coz 3*3^5=3^(5+1)=3^6
yes and thats why it is 3 * 3^5. Now 3* 3^5=3^6
Ummm I am trying to say that its a formula for exponents which can be explained in multiple ways. And I suggest you take any book even the OG and go through the exponents basics once. Should not take more than 20 minutes.
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by Ian Stewart » Thu Jul 16, 2009 1:47 pm
mehravikas wrote:Thanks..Basically I got confused in a step that was not explained in the answer explanations:

3^5+3^5+3^5 = 3^5(1 + 1 + 1) = 3^5 * 3^1 = 3^6
Nothing wrong with factoring here, as you did above, to see that 3^5 + 3^5 + 3^5 = 3^6. You can also notice that we're simply adding the same thing three times, which is what it means to multiply by 3:

x + x + x = 3x

so 3^5 + 3^5 + 3^5 = 3*(3^5)
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