What is the smallest integer \(n\) for which \(25^n > 5^{12}?\)

Gmat_mission wrote: ↑

Fri Jan 08, 2021 1:55 am

What is the smallest integer \(n\) for which \(25^n > 5^{12}?\)

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

Answer: B

Source: Official Guide

We have: 25^n > 5^12

To rewrite this inequality with the SAME base, we'll replace 25 with 5².
When we do so, we get: (5²)^n > 5^12
Apply the Power of a Power law to get: 5^(2n) > 5^12

This means that it must be the case that 2n > 12
Divide both sides of the inequality by 2 to get: n > 6

What is the smallest integer n for which 25^n > 5^12 ?
We now know that n > 6
So, 7 is the smallest possible INTEGER value that satisfies this inequality.

Answer: B

Cheers,
Brent

Gmat_mission wrote: ↑

Fri Jan 08, 2021 1:55 am

What is the smallest integer \(n\) for which \(25^n > 5^{12}?\)

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

Answer: B

Source: Official Guide

Solution:

We re-express 25 as 5^2:

25^n > 5^12

5^(2n) > 5^12

When we have the same base, we can compare the exponents:

2n > 12

n > 6

The smallest integer greater than 6 is 7.

Answer: B