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8√4n=(4√2)^8, n=?

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8√4n=(4√2)^8, n=?

by Max@Math Revolution » Sun Jul 10, 2016 11:49 pm
8√4n=(4√2)^8, n=?
A. 4^8
B. 4^10
C. 4^12
D. 4^16
E. 4^32

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by Max@Math Revolution » Tue Jul 12, 2016 5:51 pm
The correct answer is D.
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by Brent@GMATPrepNow » Sat Jul 16, 2016 2:12 pm
Max@Math Revolution wrote:If 8√(4n) = (4√2)^8, n = ?
A. 4^8
B. 4^10
C. 4^12
D. 4^16
E. 4^32
Given: 8√(4n) = (4√2)^8

Notice that 8√(4n) = 8(√4)(√n) = 8(2)(√n) = 16√n

Also, (4√2)^8 = [(4)^8][(√2)^8]
= [4^8][(√2)(√2)(√2)(√2)(√2)(√2)(√2)(√2)]
= [4^8][(2)(2)(2)(2)]
= [4^8][(4)(4)]
= [4^8][4^2]
= 4^10

So the given equation 8√(4n) = (4√2)^8...
Becomes: 16√n = 4^10
Rewrite as: (4^2)√n = 4^10
Divide both sides by 4^2 to get: √n = 4^8
Square both sides to get: n = 4^16

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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