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Exponents

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Source: — Problem Solving |

by Sakhalin » Tue May 06, 2008 7:20 am
I think the answer for n=21

5^21*4^11=2*10^n
5^21*2^22=2*5^n*2^n
Threfore n must equal 21

I am not sure. Please let me know if you know the correct solution
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by Sakhalin » Tue May 06, 2008 7:21 am
I think the answer for n=21

5^21*4^11=2*10^n
5^21*2^22=2*5^n*2^n
Threfore n must equal 21

I am not sure. Please let me know if you know the correct solution
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by VP_RedSoxFan » Tue May 06, 2008 7:28 am
The only way to solve this one is to realize what the exponents really mean.

You can think of 5^21 as a list of 21 5's multiplied together.

4^11 = (2^2)^11 = 2^22 and you can think of this as a list of 22 2's multiplied together.

Instead of multiplying the all of the 5's and then all of the 2's, then multiplying their products together, you can match up a 5 and 2. Now instead of 21 5's and 22 2's, you have 21 10's and a 2 to multiply together. And, clearly, another way to write that would be 2 * 10^21

The answer is 21
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by AleksandrM » Tue May 06, 2008 9:53 am
I got 21 as well. The key in this one is to see that when you split 10^n the result is not 2 * 5^n but is 2^n * 5^n. Another mistake that could be made is to ignore the 2 and forget that it is 2^1, leaving a 1 behind.
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