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5^21 * 4^11 = 2 * 10^N

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by ildude02 » Wed Jul 23, 2008 6:53 pm
you can reduce the equation futhur,


5 ^ 21 x 2 ^22 = 2 .(5x2)^n
=> 5 ^21 .x 2 ^ 22 = 2 ^n+1 x 5^n; equating the similar powers, n = 21.
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by pepeprepa » Wed Jul 23, 2008 9:47 pm
21 as well
My way:
5^10 x 20^11 = 20 x 10^(N-1)
(5x20)^10=10^(N-1)
10^20=10^(N-1)
N=21
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by gmatutor » Thu Jul 24, 2008 1:17 am
Prime factor
5^21 * 4^11 = 2 * 10^N =>

5^21 * (2^2)^11 = 2* (2*5)^N

There must be the same number of each prime on both sides of the equation. Since there are 21 fives on one side there must be 21 fives on the other.

N=21
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by mdwolman » Fri Jan 14, 2011 4:24 am
Hi,

How do you get from 2 x (5x2)^n to ----> 2^n+1 x 5^n?
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by arora007 » Fri Jan 14, 2011 4:35 am
mdwolman, you have to balance out the 5^21 on rhs,right?

now the 10 = (2*5) , now whatever be the power of 2 , you need the power of 5 to be balanced...

so feel confident while choosing N=21
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by Rahul@gurome » Fri Jan 14, 2011 5:38 am
mdwolman wrote:How do you get from 2 x (5x2)^n to ----> 2^n+1 x 5^n?
We can write (5*2)^n = (2^n)*(5^n)

Thus 2*(5*2)^n = 2*(2^n)*(5^n) = (2*(2^n))*(5^n) = (2^(n + 1))*(5^n)

It is an application of a property of exponents,
  • (x^a)*(x^b) = x^(a + b)
Here, x = 2, a = 1 and b = n.
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by Rahula g » Fri Jan 14, 2011 8:21 am
itchyallover wrote:
5^21 * 4^11 = 2 * 10^N
N=?
5^21 * 2^11 * 2^11 = 2 * 5^N * 2^N

5^21 * 2^22 = 5^N * 2^N+1

N= 21
rahul@gurome wrote:
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