## Large exponents

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### Large exponents

by krithika1993 » Wed Sep 07, 2016 2:29 pm
Hi there,

I was wondering if someone could help me determine an efficient approach to solving problem-solving questions that ask for what the exponent of a product would be in a mathematical equation. For instance, here is a question that illustrates my concern:

QUESTION

If 5^21 x 4^11 = 2 x 10^n, what is n?
a) 11
c) 22
d) 23
e) 32

Should I simplify 4^11 somehow/try to simplify each term to the same "base" value? Totally confused with such questions that deal with very large exponents...

Thank you
Krithika

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by [email protected] » Wed Sep 07, 2016 4:05 pm
Hi krithika1993,

Exponent questions on the GMAT come down to just a handful of rules, but those rules all involve having similar 'base' numbers.

In this question, notice how the bases on the 'left side' of the equation are 5 and 2 (4 can be rewritten as 2^2) and that on the 'right side' the bases are 2 and 5 (the 10 can be rewritten as 2x5). Considering that there's a 5^21 on the 'left side', what does the N have to equal...? (and you can confirm that answer by thinking about the 2s on both sides of the equation.

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by Cthulu » Wed Sep 07, 2016 6:18 pm
This was my way of doing it stepwise :

5^21 * 4^11

5^10 * 5^11 * 4^11

5^10 * 20^11

5^10 * 2^11 * 10^11

5^10 * 2^10 * 2^1 10^11

10^10 * 2^1 * 10^11

2^1 * 10^21

hence n is 21

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by Cthulu » Wed Sep 07, 2016 6:21 pm
Even simpler is :

5^21 * 4^11

5^11 * 5^10 * 2^11 * 2^10 * 2^1

10^11 * 10^10 * 2^1

10^21 * 2^1

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by [email protected] » Thu Sep 08, 2016 4:50 pm
krithika1993 wrote:Hi there,

I was wondering if someone could help me determine an efficient approach to solving problem-solving questions that ask for what the exponent of a product would be in a mathematical equation. For instance, here is a question that illustrates my concern:

QUESTION

If 5^21 x 4^11 = 2 x 10^n, what is n?
a) 11
c) 22
d) 23
e) 32
Here is my take on this question:

You can start by breaking each (non-prime) base into prime factors.

4^11 = (2 x 2)^11 = (2^2)^11 = 2^22

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

So our equation now looks like:

5^21 x 2^22 = 2^1 x 2^n x 5^n

5^21 x 2^22 = 2^(1+n) x 5^n

Since we have bases of 5 and 2 on both sides of our equation, we can set the bases of 2 equal to each other or the bases of 5 equal to each other and determine n.

It is easiest to equate the bases of 5.

5^21 = 5^n

21 = n

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by [email protected] » Thu Sep 15, 2016 9:21 pm
Don't trust anyone who doesn't use superscripts

5Â²Â¹ * 4Â¹Â¹ = 2 * 10â�¿

5Â²Â¹ * (2Â²)Â¹Â¹ = 2 * (2*5)â�¿

5Â²Â¹ * 2Â²Â² = 2Â¹ * 2â�¿ * 5â�¿

We have to have the same exponent on the 5s, so n = 21.

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