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ian7777
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ian7777
No more takers???

anything missed??????????/
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[email protected]
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It should be 'E'.

[2^(x+1)]^5 = sqrt[2^10x] * 2^5 = 3*2^7
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Himalayan
ian7777
No more takers???
anything missed??????????/


No, you were right. Answer's E.
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nice one vshaunak!!!

[quote="[email protected]"]It should be 'E'.

2^(x+1)]^5= sqrt[2^10x] * 2^5 = 3*2^7[/quote]
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FN
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I m trying to follow your algebra here

2^10x=(2^5x)^2=12^2, sqrt both sides..2^5x=12; got it

2^(x+1)^5 can be written as (2^x * 2^2)^5; which gives 2^5x *2^10?? this where i get confused, can someone please explain?

Himalayan
ian7777



2^(10x) = 144
(2^(5x))^2 = 12^2
so 2^5x = 12

=(2^(x+1))^5
=(2^x 2)^5
= (2^5x) (2^5)
= 12 (2^5)
= 3 (2^7)
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ian7777
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fresinha12
I m trying to follow your algebra here

2^10x=(2^5x)^2=12^2, sqrt both sides..2^5x=12; got it

2^(x+1)^5 can be written as (2^x * 2^2)^5; which gives 2^5x *2^10?? this where i get confused, can someone please explain?



No, you made one mistake.

2^(x+1)^5 can be written as
2^(5x + 5)

which can be written as
2^5x * 2^5

And 2^5 is 32

and 2^5x is 12

so it's 12*32

and 12 reduces to 3*2*2, which is 3*2^2
and the 32 is 2^5
So all together it's

3*2^7
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dang..i dont know where my mind is these days...

thanks ian7777 ;)

ian7777
fresinha12
I m trying to follow your algebra here

2^10x=(2^5x)^2=12^2, sqrt both sides..2^5x=12; got it

2^(x+1)^5 can be written as (2^x * 2^2)^5; which gives 2^5x *2^10?? this where i get confused, can someone please explain?


No, you made one mistake.

2^(x+1)^5 can be written as
2^(5x + 5)

which can be written as
2^5x * 2^5

And 2^5 is 32

and 2^5x is 12

so it's 12*32

and 12 reduces to 3*2*2, which is 3*2^2
and the 32 is 2^5
So all together it's

3*2^7



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