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Help with this problem please

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by VP_RedSoxFan » Thu Jul 31, 2008 8:26 am
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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by Ilikemeat321 » Thu Jul 31, 2008 4:28 pm
thanks ! go redsox too bad they got ride of manny
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