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# If n is a number such that (−8)^(2n) = 2^(8+2n), then n =

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Joined: 02 Sep 2009
Posts: 54369
If n is a number such that (−8)^(2n) = 2^(8+2n), then n =  [#permalink]

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01 Jun 2017, 13:09
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81% (01:17) correct 19% (02:04) wrong based on 140 sessions

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If n is a number such that $$(−8)^{(2n)} = 2^{(8+2n)}$$, then n =

(A) 1/2
(B) 3/2
(C) 2
(D) 4
(E) 5

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Re: If n is a number such that (−8)^(2n) = 2^(8+2n), then n =  [#permalink]

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01 Jun 2017, 13:39
(−8)^(2n)=2^(8+2n)
(-2)^6n= 2^(8+2n)
If positive, negative does not matter
6n = 8+2n
n=4
D
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If n is a number such that (−8)^(2n) = 2^(8+2n), then n =  [#permalink]

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01 Jun 2017, 15:58
1
1
Bunuel wrote:
If n is a number such that $$(−8)^{(2n)} = 2^{(8+2n)}$$, then n =

(A) 1/2
(B) 3/2
(C) 2
(D) 4
(E) 5

$$(−8)^{(2n)} = 2^{(8+2n)}$$
Any number with negative raised to even number exponent will always be positive. Therefore $$(−8)^{(2n)}$$ can be written as $$(8)^{(2n)}$$

$$(8)^{(2n)} = 2^{(8+2n)}$$

$$2^3^{(2n)} = 2^{(8+2n)}$$

$$2^{6n} = 2^{(8+2n)}$$

Therefore;
6n = 8 + 2n
6n - 2n = 8
4n = 8
n = $$\frac{8}{4}$$ = 2
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Re: If n is a number such that (−8)^(2n) = 2^(8+2n), then n =  [#permalink]

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02 Jun 2017, 00:02
2
amastu wrote:
(−8)^(2n)=2^(8+2n)
(-2)^6n= 2^(8+2n)
If positive, negative does not matter
6n = 8+2n
n=4
D

Hi

When you wrote 6n=8+2n, you will get
6n-2n = 8 or 4n = 8 or n=2.

Your approach is correct, probably there was a typo so I am pointing that out.
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Re: If n is a number such that (−8)^(2n) = 2^(8+2n), then n =  [#permalink]

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02 Jun 2017, 09:39
1
Bunuel wrote:
If n is a number such that $$(−8)^{(2n)} = 2^{(8+2n)}$$, then n =

(A) 1/2
(B) 3/2
(C) 2
(D) 4
(E) 5

$$(−8)^{(2n)} = 2^{(8+2n)}$$

Or, $$(−2)^{(3*2n)} = 2^{(8+2n)}$$

Or, $${(6n)} = {(8+2n)}$$

Or, $$6n = 8 + 2n$$

Or, $$4n = 8$$

Or, $$n = 2$$

So, Answer must be (C) 2
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Re: If n is a number such that (−8)^(2n) = 2^(8+2n), then n =  [#permalink]

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05 Jun 2017, 15:50
1
Bunuel wrote:
If n is a number such that $$(−8)^{(2n)} = 2^{(8+2n)}$$, then n =

(A) 1/2
(B) 3/2
(C) 2
(D) 4
(E) 5

The negative sign in front of the 8 will be eliminated because we are raising a negative number (-8) to an even power. Note that (-8)^(2n) = [(-8)^2]^n = [8^2]^n = 8^(2n). Then:

8^(2n) = (2^3)^(2n) = 2^(6n) = 2^(8 + 2n)

Since bases are the same, we can equate the exponents:

6n = 8 + 2n

4n = 8

n = 2

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Re: If n is a number such that (−8)^(2n) = 2^(8+2n), then n =  [#permalink]

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10 Jun 2017, 06:37
I guess the only thing which is confusing for student here is
(−8)^(2n)=2^(8+2n)
On left there is negative sign but keep in mind since n will need to be multiplied with even number so result is always even which in turn make bracket value positive
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Re: If n is a number such that (−8)^(2n) = 2^(8+2n), then n =  [#permalink]

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16 Jul 2017, 22:37
(−8)^(2n) will always result into positive number. Hence correct option is C.
Re: If n is a number such that (−8)^(2n) = 2^(8+2n), then n =   [#permalink] 16 Jul 2017, 22:37
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