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crazy exponent problem

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Intern
Joined: 31 Jul 2008
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31 Jul 2008, 13:13
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

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Joined: 07 May 2008
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31 Jul 2008, 13:15
I like this problem a lot. Hopefully I can help.

If you change the 4^11 into (2^2)^11, then it becomes 2^22.

Now, if you look at the left side, you have 21 5's being multiplied by 22 2's. If you match up a 5 and 2, you'll get 21 pairs of 5*2 with one 2 left over. Turn those 21 pairs of a 5 and 2 into 10 and you're left with 21 10's multiplied together and then multiplied by 2.

That equals 2* (10^21) so N=21.

Algebraically, (ab)^c = a^c * b^c as a rule of exponents. After changing 4^11 into 2^22 you can apply this rule backwards as 5^21 * 2^21 * 2 = (5*2)^21 * 2 = 2* (10^21), N = 21
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31 Jul 2008, 13:19

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

5^21 * 4^11 = 2 * 10^N
-->
5^21 * (2*2)^11 = 2 * 10^N
-->
5^21 * (2)^22 = 2 * 10^N
5^21 * (2)^21 = 10^N
(5*2)^21 = 10^N
(10)^21 = 10^N
-->N=21
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31 Jul 2008, 13:23
this one has been discussed quite frequently
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03 Aug 2008, 06:14

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

its just simplifying the problm is what is important here :
5^21 * 4^11 = 5^21 * 2^22 = (10^21) *2 =2*10^N => N=21

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Re: crazy exponent problem   [#permalink] 03 Aug 2008, 06:14
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