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math question (anyone know how to solve)

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by vasbli » Sat Jun 20, 2009 2:03 pm
(1/5)^m * (1/4)^18 = 1/2(10)^35
--------
(1/5)^m * (1/4)^18=(1/5)^m * (1/2)^36=(1/5)^m * 1/(2*2^35)
and
1/2(10)^35=1/(2*5^35*2^35)

m=35
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THANK YOU

by LSJK » Sun Jun 21, 2009 1:24 pm
thank you.
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by BlindVision » Sun Jun 21, 2009 3:18 pm
vasbli wrote:(1/5)^m * (1/4)^18 = 1/2(10)^35
--------
(1/5)^m * (1/4)^18=(1/5)^m * (1/2)^36=(1/5)^m * 1/(2*2^35)
and
1/2(10)^35=1/(2*5^35*2^35)

m=35
Can you please tell me how...
(1/4)^18 became (1/2)^36?

And...

How 1/2(10)^35=1/(2*5^35*2^35) ? Thank you!
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by ssmiles08 » Sun Jun 21, 2009 3:27 pm
BlindVision wrote:
vasbli wrote:(1/5)^m * (1/4)^18 = 1/2(10)^35
--------
(1/5)^m * (1/4)^18=(1/5)^m * (1/2)^36=(1/5)^m * 1/(2*2^35)
and
1/2(10)^35=1/(2*5^35*2^35)

m=35
Can you please tell me how...
(1/4)^18 became (1/2)^36?

And...

How 1/2(10)^35=1/(2*5^35*2^35) ? Thank you!
BlindVision,

1/4 = (1/2)^2
so [(1/2)^2]^18 = (1/2)^36


and

1/[2*10] = 1/[2*5*2] = 1/[(2*5)*(2)]
since the whole bottom of the fraction is raised to the power of 35, you need to distribute that property to all multiplied products.

= 1/(2*5^35*2^35)

-hope that helps.
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by LSJK » Sun Jun 21, 2009 3:30 pm
Question 1:

4= 2^2

so (2^2)^18= 2^36 because 2 * 18=36

______________________________________
Question 2:

2^2 * 2= 2^3=8

so

2(10)^35 = 2 * 2^35 * 5^35

so based on above. . .

2 * 2^35 = 2^36
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by BlindVision » Sun Jun 21, 2009 3:41 pm
Aha! I get it! Thank you, SSmiles!

Thanks LSJK, I see your reply in the "post a reply" screen, but not in the regular screen.
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