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

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EgmatQuantExpert
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If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement [#permalink] New post  Updated on: 20 Mar 2017, 06:06
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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


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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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niks18
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If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement [#permalink] New post  21 Oct 2017, 00:56
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Mahmud6 wrote:
7 months have already been passed. Would you please provide OE?


Hi Mahmud6

The answer would be C

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

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

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] New post  20 Mar 2017, 05:31
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Re: If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement [#permalink] New post  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] New post  21 Oct 2017, 01:54
niks18 wrote:
Mahmud6 wrote:
7 months have already been passed. Would you please provide OE?


Hi Mahmud6

The answer would be C

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

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

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,

Could you please help me understand how 65.2^{(x+1)} is converted to 65a? Isn't 65.2^{(x+1)} = (65+0.2)^{(x+1)}?

Thank you.

Paul
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niks18
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Re: If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement [#permalink] New post  21 Oct 2017, 01:57
Paulli1982 wrote:
niks18 wrote:
Mahmud6 wrote:
7 months have already been passed. Would you please provide OE?


Hi Mahmud6

The answer would be C

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

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

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,

Could you please help me understand how 65.2^{(x+1)} is converted to 65a? Isn't65.2^{(x+1)} = (65+0.2)^{(x+1)}?

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] New post  21 Oct 2017, 07:47
niks18 wrote:
Mahmud6 wrote:
7 months have already been passed. Would you please provide OE?


Hi Mahmud6

The answer would be C

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

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

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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niks18
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If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement [#permalink] New post  21 Oct 2017, 07:52
rahul16singh28 wrote:
niks18 wrote:
Mahmud6 wrote:
7 months have already been passed. Would you please provide OE?


Hi Mahmud6

The answer would be C

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

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

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 \(2^x\)
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Re: If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement [#permalink] New post  22 Aug 2021, 12:01
niks18 wrote:
Mahmud6 wrote:
7 months have already been passed. Would you please provide OE?


Hi Mahmud6

The answer would be C

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

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

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



I dont exactly understand how 2^{(2x+2)}.2^3-65.2^{(x+1)}+8=0 => 2^{2(x+1)}.2^3-65.2^{(x+1)}+8=0 is in the form of a^2 - 2ab + b^2.

Instead

Isolate the 2^x and take 2^x as a

2^x * 32 - 65 * 2^x*2 + 8 = 0

Remove 2 as a common factor.

Then post factorization we get 16a = 1 or a=4
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Re: If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement [#permalink] New post  22 Apr 2023, 11:31
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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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