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If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement

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If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement  [#permalink]

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Updated on: 20 Mar 2017, 06:06
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36% (01:36) correct 64% (01:50) wrong based on 83 sessions

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If $$2^{2x+5} - 65.2^{x+1} = - 8$$, then which of the following statement is definitely correct?

A. x can only be a positive integer
B. x can only be a negative integer
C. x can be either a positive integer or a negative integer
D. No value of x exists which satisfies the equation
E. x is not an integer

Thanks,
Saquib
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Originally posted by EgmatQuantExpert on 20 Mar 2017, 05:30.
Last edited by Bunuel on 20 Mar 2017, 06:06, edited 1 time in total.
Renamed the topic and edited the question.
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If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement  [#permalink]

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21 Oct 2017, 00:56
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Mahmud6 wrote:

Hi Mahmud6

$$2^{(2x+5)} - 65.2^{(x+1)}=-8$$

$$2^{(2x+2)}.2^3-65.2^{(x+1)}+8=0 => 2^{2(x+1)}.2^3-65.2^{(x+1)}+8=0$$

Let $$2^{(x+1)}=a$$, so we have

$$8a^2-65a+8=0$$ or

$$(8a-1)(a-8)=0 => a=\frac{1}{8}$$ or $$a=8$$

so $$2^{(x+1)}=\frac{1}{8}=2^{-3} => x=-4$$ (negative)

or $$2^{(x+1)}=8=2^3 =>x=2$$ (positive)

Option C
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Re: If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement  [#permalink]

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20 Mar 2017, 05:31
Reserving this space to post the official solution.
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Re: If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement  [#permalink]

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21 Oct 2017, 00:09
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Re: If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement  [#permalink]

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21 Oct 2017, 01:54
niks18 wrote:
Mahmud6 wrote:

Hi Mahmud6

$$2^{(2x+5)} - 65.2^{(x+1)}=-8$$

$$2^{(2x+2)}.2^3-65.2^{(x+1)}+8=0 => 2^{2(x+1)}.2^3-65.2^{(x+1)}+8=0$$

Let $$2^{(x+1)}=a$$, so we have

$$8a^2-65a+8=0$$ or

$$(8a-1)(a-8)=0 => a=\frac{1}{8}$$ or $$a=8$$

so $$2^{(x+1)}=\frac{1}{8}=2^{-3} => x=-4$$ (negative)

or $$2^{(x+1)}=8=2^3 =>x=2$$ (positive)

Option C

Hi,

Thank you.

Paul
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Re: If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement  [#permalink]

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21 Oct 2017, 01:57
Paulli1982 wrote:
niks18 wrote:
Mahmud6 wrote:

Hi Mahmud6

$$2^{(2x+5)} - 65.2^{(x+1)}=-8$$

$$2^{(2x+2)}.2^3-65.2^{(x+1)}+8=0 => 2^{2(x+1)}.2^3-65.2^{(x+1)}+8=0$$

Let $$2^{(x+1)}=a$$, so we have

$$8a^2-65a+8=0$$ or

$$(8a-1)(a-8)=0 => a=\frac{1}{8}$$ or $$a=8$$

so $$2^{(x+1)}=\frac{1}{8}=2^{-3} => x=-4$$ (negative)

or $$2^{(x+1)}=8=2^3 =>x=2$$ (positive)

Option C

Hi,

Thank you.

Paul

Hi Paulli1982

Dot in algebra mean multiplication so it is $$65*2^{x+1}$$ and not a decimal number
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Re: If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement  [#permalink]

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21 Oct 2017, 07:47
niks18 wrote:
Mahmud6 wrote:

Hi Mahmud6

$$2^{(2x+5)} - 65.2^{(x+1)}=-8$$

$$2^{(2x+2)}.2^3-65.2^{(x+1)}+8=0 => 2^{2(x+1)}.2^3-65.2^{(x+1)}+8=0$$

Let $$2^{(x+1)}=a$$, so we have

$$8a^2-65a+8=0$$ or

$$(8a-1)(a-8)=0 => a=\frac{1}{8}$$ or $$a=8$$

so $$2^{(x+1)}=\frac{1}{8}=2^{-3} => x=-4$$ (negative)

or $$2^{(x+1)}=8=2^3 =>x=2$$ (positive)

Option C

Hi,

May I please know what is wrong with the below approach:

2^x * 32 - 65*2*2^x = -8
2^x = (2/7)^2

Please let me know where I am wrong in this.
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If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement  [#permalink]

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21 Oct 2017, 07:52
rahul16singh28 wrote:
niks18 wrote:
Mahmud6 wrote:

Hi Mahmud6

$$2^{(2x+5)} - 65.2^{(x+1)}=-8$$

$$2^{(2x+2)}.2^3-65.2^{(x+1)}+8=0 => 2^{2(x+1)}.2^3-65.2^{(x+1)}+8=0$$

Let $$2^{(x+1)}=a$$, so we have

$$8a^2-65a+8=0$$ or

$$(8a-1)(a-8)=0 => a=\frac{1}{8}$$ or $$a=8$$

so $$2^{(x+1)}=\frac{1}{8}=2^{-3} => x=-4$$ (negative)

or $$2^{(x+1)}=8=2^3 =>x=2$$ (positive)

Option C

Hi,

May I please know what is wrong with the below approach:

2^x* 32 - 65*2*2^x = -8
2^x = (2/7)^2

Please let me know where I am wrong in this.

hi rahul16singh28

the highlighted portion is not correct. it is $$2^{2x}$$ and not $$2^x$$
If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement &nbs [#permalink] 21 Oct 2017, 07:52
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