## OG -106

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### OG -106

by neeg » Sat Mar 30, 2013 11:43 am

00:00

A

B

C

D

E

## Global Stats

What is the smallest integer n for which 25^n > 5^12?

a)6
b)7
c)8
d)9
e)10

If I try solving this by rewriting the right side rather than the left side as a common base i.e. 5 ^12 --> 25^ 11 then n would have to be greater than 11 ?

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by [email protected] » Sat Mar 30, 2013 12:04 pm
neeg wrote:What is the smallest integer n for which 25^n > 5^12?

a)6
b)7
c)8
d)9
e)10
First recognize that 5^12 = (5^2)^6 = 25^6
What is the smallest integer n for which 25^n > 25^6?
We need n>6
So, n = 7 is the smallest integer.

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by rintoo22 » Sat Mar 30, 2013 12:05 pm
25^n > 5^12 => 5^2n > 5^12

Therefore 2n=12 => n=6. So for n=6 both sides will match.

But the question asks the smallest integer n for which 25^n > 5^12. So the the left hand side shodl be greater. So the answer in my opinion should be b.7.

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by [email protected] » Sat Mar 30, 2013 10:33 pm
neeg wrote:If I try solving this by rewriting the right side rather than the left side as a common base i.e. 5 ^12 --> 25^ 11 then n would have to be greater than 11 ?
Arithmetic of powers do not work that way.
5^12 = 5*(5^11) NOT 25^11
Or, 5^12 = (5^6)*(5^6) = (5*5)^6 = 25^6

To clear your confusion, let us go back to the very definition of powers,
a^n = a multiplied with it self n times = a*a*a*... (n times)
a^m = a multiplied with it self m times = a*a*a*... (m times)

Now,
• a^(m + n) = (a^m)*(a^n)
a^(mn) = (a^m)^n = (a^n)^m
Hope that helps.
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by neeg » Sun Mar 31, 2013 5:27 pm
Brilliant as always. I can see the error, Thank you.
[email protected] wrote:
neeg wrote:If I try solving this by rewriting the right side rather than the left side as a common base i.e. 5 ^12 --> 25^ 11 then n would have to be greater than 11 ?
Arithmetic of powers do not work that way.
5^12 = 5*(5^11) NOT 25^11
Or, 5^12 = (5^6)*(5^6) = (5*5)^6 = 25^6

To clear your confusion, let us go back to the very definition of powers,
a^n = a multiplied with it self n times = a*a*a*... (n times)
a^m = a multiplied with it self m times = a*a*a*... (m times)

Now,
• a^(m + n) = (a^m)*(a^n)
a^(mn) = (a^m)^n = (a^n)^m
Hope that helps.

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by GMATGuruNY » Sun Mar 31, 2013 7:34 pm
neeg wrote:What is the smallest integer n for which 25^n > 5^12?

a)6
b)7
c)8
d)9
e)10

If I try solving this by rewriting the right side rather than the left side as a common base i.e. 5 ^12 --> 25^ 11 then n would have to be greater than 11 ?
An alternate approach would be to plug in the answers, starting with the smallest answer choice.

If n=6, we get:
25â�¶ > 5Â¹Â²
(5Â²)â�¶ > 5Â¹Â²
5Â¹Â² > 5Â¹Â².
Since n=6 makes the lefthand side EQUAL to the righthand side, the smallest possible value of n must be 7.

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### Re: OG -106

by [email protected] » Fri Apr 23, 2021 2:53 pm
Hi All,

We’re asked for the SMALLEST possible INTEGER value for N that will make 25^N > 5^12. While the answers to this question are numbers (so we could TEST THE ANSWERS), this prompt is based around some standard Exponent rules, so approaching it with Arithmetic should be fairly quick.

The “base” of an exponent calculation can sometimes be ‘re-written’ (if the base has any factors that are greater than 1, then you can rewrite the calculation while keeping its value the same). Here, 25 can be re-written as 5^2, so the “left side” of the inequality can be re-written as (5^2)^N.

When ‘raising a power to a power’, you multiply the Exponents, meaning that we now have:

5^(2N) > 5^12

Thus, we need 2N > 12…. N > 6. The smallest answer that fits that description is…