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(5^11) . 4^10 = 2 . 10^n

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by scoobydooby » Sun May 03, 2009 11:13 am
(5^11) . (4^20) = 2 . (10^n)
=>(5^11)*(2^2)^20=2*(2*5)^n
=>(5^11)*(2)^40=[2^(n+1)]*5^n
equating the bases on both sides
we get 2 values of n
n=11; n+1=40=>n=39. which is weird.

is there any typo in the question?
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by keizer Soze » Sun May 03, 2009 11:33 am
I solved it this way:

5^11 . 4^10 = 2 . 10^n

5^11 . 2^19 = 10^n

10^11 . 2^8 = 10^n

10^11 . 2^8 . 5^8/5^8 = 10^n

10^19 . 1/5^8 = 10^n (A)

Here: 5^8= 10^8 . 1/2^8 (B)

2^10=~ 10^3............ so, 2^8 =~ 10^2,4

Again in (B): 5^8 = 10^8 / 10^2,4 = 10^5,6

Finally in (A): 10^19 / 10^5,6 =~ 10^n

10^13,4 =~ 10^n

n=~ 13,4

I think that this steps are Ok....
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by keizer Soze » Sun May 03, 2009 11:35 am
Sorry I realized that there was one... 4^10 and not 4^20...
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by PAB2706 » Sun May 03, 2009 10:46 pm
i solved in the following way.


5^11 . 4^10 = 2 . 10^n

LHS goes..

5^11 . 2^20

2* (5^11*2^19)

2* (5^11*2^11) * 2^8

2* (10^11) * 2^8 -------I

now 2^8 =256

thus more than 10^2 but less than 10^3

thus we may consider 10^2 <2^8 < 10^3

thus n is between 10^11*10^2 and 10^11*10^3 ie 13 < n <14

hope this helps
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