If 5^21 x 4^11 = 2 x 10^n what is the value of n?

[agha79]

I was unable to find below question in the form. I got this question during my GMAT prep exam one. I am not even sure how to get started on this question.

If 5^21 x 4^11 = 2 x 10^n what is the value of n?
A) 11
B) 21
C) 22
D) 23
E) 32

OA: B

[jnelson0612]

I was unable to find below question in the form. I got this question during my GMAT prep exam one. I am not even sure how to get started on this question.

If 5^21 x 4^11 = 2 x 10^n what is the value of n?
A) 11
B) 21
C) 22
D) 23
E) 32

OA: B

First thing to do here is to rewrite the left side of the question by breaking that 4^11 down to a prime factorization. 4^11= (2^2)^11, or 2^22.

Okay, so now we have 5^21 * 2^22. Notice how the right side has 10^n. Maybe we should try to make some 10s on the left side. What if we rewrite the left side as this: 5^21 * 2^21 * 2^1? Now we can combine 5^21 * 2^21 and make 10^21 (plus let's not forget the leftover 2^1, or just 2).

Notice what we have now: 10^21 * 2 = 2 * 10^n. n must be 21, or B.

[agha79]

Hi Jamie -
Thanks again. I always find exponent problems very intimidating. It seems that backing down the problem is key here. In future as well do you recommend for similar problems I should try to brake down the problem to an extent so it matches the side where the variable is? Or this approach is very specific to this problem.

[jnelson0612]

Hi Jamie -
Thanks again. I always find exponent problems very intimidating. It seems that backing down the problem is key here. In future as well do you recommend for similar problems I should try to brake down the problem to an extent so it matches the side where the variable is? Or this approach is very specific to this problem.

Hi Agha,
Good observation! Yes, you normally want to break down the problem so that you get some sort of match between the sides. Glad this helped.