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by MarkIppo » Tue Apr 03, 2007 12:40 pm
This one is probably easy for you guys, but it is stumping me.

5^21 x 4^11=2 x 10^n

Reads: 5 to the 21st power mult. by 4 to the 11th power equals 2 mult by 10 to the nth power. What is n?

I know the answer because I took the practice test, but how does one get there?
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by gmatme » Tue Apr 03, 2007 1:42 pm
5^21 * 4 ^11 = 2 * 10^n
5 ^21 * (2^2)^11 = 5 ^21 * 2 ^22 = 5 ^21 * 2 ^21 * 2 = 10 ^21 * 2

So n = 21.
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by vk.neni » Tue Apr 03, 2007 1:43 pm
Reorganizing the following equation a little differently leads us to the answer quite quickly.

5^21 x 4^11=2 x 10^n

5^21 x (2^2)^11 = 2 x 10^n

5^21 x 2^22 = 2 x 10^n

(5x2)^21 x 2 = 2 x 10^n

10^21 = 10^n


So n=21.


-Neni
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by Cybermusings » Wed Apr 04, 2007 3:57 am
5^21 x 4^11=2 x 10^n

5^21 x (2^2)^11= 2 x 10^n

5^21 x 2^22 = 2 x 10^n

5^21 x 2^21 x 2 = 2 x 10^n

10^21 x 2 = 2 x 10^n

10^21 = 10^n

Thus n = 21
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by vinviper1 » Mon May 12, 2008 10:34 am
(2^2)^11 = 2^22

Is this a rule of some sort that can be used for all exponents? Thanks.
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by II » Mon May 12, 2008 12:08 pm
vinviper1 wrote:(2^2)^11 = 2^22

Is this a rule of some sort that can be used for all exponents? Thanks.
We notice that 4 is a perfect square (i.e. 2^2). So we can rewrite "4^11" as (2^2)^11 ... which is simplified to 2^22.
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by Stuart@KaplanGMAT » Mon May 12, 2008 1:19 pm
vinviper1 wrote:(2^2)^11 = 2^22

Is this a rule of some sort that can be used for all exponents? Thanks.
Yes:

(x^a)^b = x^(a*b)

e.g.:

(2^5)^3 = 2^(5*3) = 2^15

Similarly:

(x^a) * (x^b) = x^(a+b)

e.g."

(2^5) * (2^3) = 2^(5+3) = 2^8

And:

(x^a) / (x^b) = x^(a-b)

(2^5) / (2^3) = 2^(5-3) = 2^2
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by vinviper1 » Wed May 14, 2008 8:03 am
Thanks Stuart!
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