Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: A very difficult math [#permalink]
23 Aug 2011, 17:57

Jasonammex wrote:

If x and K are integers and (12^x)(4^2x+1)=(2^k)(3^2), what is the value of K? (12^x)(4^(2x+1))=(2^k)(3^2) ==> (3^x)(4^x)(4^(2x+1))=(2^k)(3^2) ==> (3^x)(4^(3x+1))=(3^2)(4^(k/2))

So 3^x = 3^2 ==>x = 2 4^(3x+1) = 4^(k/2) ==> k = 2(3x+1) = 2(3*2+1) = 14 So OA -E a) 5 b)7 c)10 d)12 e)14

Pls explain me and there gotta be a simple way to solve this?

Re: A very difficult math [#permalink]
23 Aug 2011, 17:59

(12^x)(4^(2x+1)) = (2^k)(3^2)

we want to get the bases to match, so we can then solve for k in the exponent.

12 can be broken down as 3 x 2 x 2. As the rules for exponents go, if the exponents are the same then you can perform the operation as normal with the bases so 12^x = (2^x)(2^x)(3^x). you also want the 4 down to a base of 2, and square root of 4 is 2 so 2^2=4, so again using another exponent rule : (2^2)^(2x+1) we multiply the exponents, and get 2^(4x+2)

(2^2x)(3^x)(2^(4x+2)) = (2^k)(3^2)

Now we've got two terms with a base of 2, so we multiply them : which means we add the exponents:

(2^(2x+4x+2))(3^x) = (2^k)(3^2)

Now we can compare both sides, since the bases are the same. So we can assume, since 3^x = 3^2, then x=2 and 2^(2x+4x+2) = 2^k, therefore 2x+4x+2 = k

Substituting 2 for x, we get

2(2) + 4(2) + 2 = k 4 + 8 + 2 = k 14 = k

Answer : E

Last edited by meshell on 23 Aug 2011, 18:55, edited 1 time in total.

For my Cambridge essay I have to write down by short and long term career objectives as a part of the personal statement. Easy enough I said, done it...