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If 2^(2x) + 2^(x)  6 = 0, x  1/x = ? [#permalink]
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09 Sep 2017, 08:10
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Re: If 2^(2x) + 2^(x)  6 = 0, x  1/x = ? [#permalink]
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09 Sep 2017, 08:29
It's an E. The question breaks down to 1/(4^x)+1/(2^x)6=0. Let 1/2^x=a. Then it becomes a^2+a6=0. On solving we get a=2 and 3. Thus 1/(2^x)=2 Which gives x=1. So x1/x equals 1+1=2. My formatting may not be very good but have tried to keep it lucid. Sent from my Lenovo TAB S850LC using GMAT Club Forum mobile app



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Re: If 2^(2x) + 2^(x)  6 = 0, x  1/x = ? [#permalink]
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09 Sep 2017, 08:44
Since we have been asked to find the value of \(x  \frac{1}{x}\), this can be rewritten as \(\frac{x^2  1}{x}\) The equation \(2^{(2x)} + 2^{(x)}  6 = 0\) => \(\frac{1}{{2^{(x)^2}}} + \frac{1}{{2^{(x)}}}  6 = 0\) Substituting \(\frac{1}{2^x}\) be b Therefore, \(b^2 + b  6 = 0\) Solving for b, b=3 or 2 If b=2  \(\frac{1}{2^x} = 2^{x} = 2\) So, x = 1 The value for the expression \(\frac{x^2  1}{x} = \frac{0}{1} = 0\) (Option C)
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If 2^(2x) + 2^(x)  6 = 0, x  1/x = ? [#permalink]
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09 Sep 2017, 08:47
kumarparitosh123 wrote: It's an E. The question breaks down to 1/(4^x)+1/(2^x)6=0. Let 1/2^x=a. Then it becomes a^2+a6=0. On solving we get a=2 and 3. Thus 1/(2^x)=2 Which gives x=1. So x1/x equals 1+1=2. My formatting may not be very good but have tried to keep it lucid. Sent from my Lenovo TAB S850LC using GMAT Club Forum mobile appIt must be C, kumarparitosh123As x=1, and we need to find the value of x1/x The expression \(x\frac{1}{x} = 1 (\frac{1}{1}) = 1 + 1 = 0\)
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Re: If 2^(2x) + 2^(x)  6 = 0, x  1/x = ? [#permalink]
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09 Sep 2017, 08:49
pushpitkc wrote: kumarparitosh123 wrote: It's an E. The question breaks down to 1/(4^x)+1/(2^x)6=0. Let 1/2^x=a. Then it becomes a^2+a6=0. On solving we get a=2 and 3. Thus 1/(2^x)=2 Which gives x=1. So x1/x equals 1+1=2. My formatting may not be very good but have tried to keep it lucid. Sent from my Lenovo TAB S850LC using GMAT Club Forum mobile appIt must be C, kumarparitosh123As x=1, and we need to find the value of x1/x The expression \(x\frac{1}{x} = 1 (\frac{1}{1}) = 1 + 1 = 0\) Yes Pushpitkc My Bad. I don't know in which state of mind did I solve.. Thanks for pointing it out. A lesson for me not to rush much..? Sent from my Lenovo TAB S850LC using GMAT Club Forum mobile app



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If 2^(2x) + 2^(x)  6 = 0, x  1/x = ? [#permalink]
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09 Sep 2017, 08:51
Bunuel wrote: If \(2^{(2x)} + 2^{(x)}  6 = 0\), \(x  \frac{1}{x} =\) ?
(A) 2 (B) 1 (C) 0 (D) 1 (E) 2 let \(2^{x} = a\) so the question becomes \(a^2 + a  6 = 0\). Solving this we get \((a2)(a+3) = 0\). Therefore \(a = 2\) or \(a =3\). But as "\(a\)" is a positive number (\(2\) raised to some power will always be positive), hence \(a = 3\) is not possible So we get \(2^{x} = 2\) or \(x = 1\) therefore \(x \frac{1}{x} = 1 +1 = 0\) Option C



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Re: If 2^(2x) + 2^(x)  6 = 0, x  1/x = ? [#permalink]
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09 Sep 2017, 12:55
Bunuel wrote: If \(2^{(2x)} + 2^{(x)}  6 = 0\), \(x  \frac{1}{x} =\) ?
(A) 2 (B) 1 (C) 0 (D) 1 (E) 2 2^x=a equation becomes a^2+a6=0...comes out to be a=3(not possible,since 2^x is not equal to 3).Thus a=2, 2^x=2 x=1 and finally the answer comes out to be 0



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Re: If 2^(2x) + 2^(x)  6 = 0, x  1/x = ? [#permalink]
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09 Sep 2017, 21:06
Bunuel wrote: If \(2^{(2x)} + 2^{(x)}  6 = 0\), \(x  \frac{1}{x} =\) ?
(A) 2 (B) 1 (C) 0 (D) 1 (E) 2 \(2^{(2x)} + 2^{(x)}  6 = 0\) => \(1/2^{(2x)} + 1/2^{(x)}  6 = 0\) Let 2^x = y => \(1/y^2 + 1/y  6 = 0\) => 6y^2  6y 1 = 0 => 6y^2  3y + 2y 1 = 0 => 3y (2y1) + 2y1 = 0 => (3y +1) (2y1) = 0 => y = 1/3, 1/2 y = 2^1 x = 1 x1/x = 1  (1/1) = 1 +1 = 0 Answer C
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Re: If 2^(2x) + 2^(x)  6 = 0, x  1/x = ? [#permalink]
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24 Sep 2017, 21:39
Plugin values of X 2ˆ(2x)+2ˆ(x)−6=0 We need a value of x which results the LHS of the equation to be zero. Trial and error we can take x=(1) I checked from given answer choices. 2ˆ(2*1)+2(*1)6=0 2ˆ2+26 4+26=0 x1/X=11/(1)=1+1=0 Answer c



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Re: If 2^(2x) + 2^(x)  6 = 0, x  1/x = ? [#permalink]
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26 Sep 2017, 16:28
Bunuel wrote: If \(2^{(2x)} + 2^{(x)}  6 = 0\), \(x  \frac{1}{x} =\) ?
(A) 2 (B) 1 (C) 0 (D) 1 (E) 2 We can let y = 2^(x) and rewrite the equation as: (2^(x))^2 + 2^(x)  6 = 0 y^2 + y  6 = 0 (y + 3)(y  2) = 0 y = 3 or y = 2 Since y = 2^(x), we can say 2^(x) = 3 or 2^(x) = 2. However, since 2 is positive, 2^(x) will be positive regardless of the value of x. So, 2^(x) can’t be 3, and thus it must be 2. Let’s solve the equation 2^(x) = 2: 2^(x) = 2^1 x = 1 x = 1 Thus, x  1/x = 1  1/(1) = 1 + 1 = 0. Answer: C
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Re: If 2^(2x) + 2^(x)  6 = 0, x  1/x = ?
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