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Large exponents

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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
b) 21 (answer)
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...


Would appreciate any help you have to offer!

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 Scott@TargetTestPrep » 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
b) 21 (answer)
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

Answer: B

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by Matt@VeritasPrep » Thu Sep 15, 2016 9:21 pm
Don't trust anyone who doesn't use superscripts :)

5²¹ * 4¹¹ = 2 * 10�

5²¹ * (2²)¹¹ = 2 * (2*5)�

5²¹ * 2²² = 2¹ * 2� * 5�

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