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
Large exponents
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- krithika1993
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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.
GMAT assassins aren't born, they're made,
Rich
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.
GMAT assassins aren't born, they're made,
Rich
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Here is my take on this question: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
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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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.

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.