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Given: 4^(2t+1)  4^(t+2) = 128, find t [#permalink]
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Updated on: 15 Nov 2013, 01:53
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Given: 4^(2t+1)  4^(t+2) = 128 Find: t
Originally posted by NvrEvrGvUp on 15 Nov 2013, 01:52.
Last edited by Bunuel on 15 Nov 2013, 01:53, edited 1 time in total.
Edited the question.



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Re: Given: 4^(2t+1)  4^(t+2) = 128, find t [#permalink]
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15 Nov 2013, 02:00
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Re: Given: 4^(2t+1)  4^(t+2) = 128, find t [#permalink]
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15 Nov 2013, 02:02
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Bunuel wrote: NvrEvrGvUp wrote: Given: 4^(2t+1)  4^(t+2) = 128 Find: t \(4^{2t+1}  4^{t+2} = 128\); \(4*4^{2t}  4^2*4^{t} = 128\); \(4*(4^t)^2 16*4^t  128=0\); \((4^t)^2 4*4^t  32=0\); Solve for 4^t: \(4^t=4\) (discard, because 4^t cannot be negative) or \(4^t=8\). \(4^t=8\) > \(2^{2t}=2^3\) > \(2t=3\) > \(t=1.5\). Hope it's clear. Hidden quadratic. Got it, thanks Bunuel.



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Re: Given: 4^(2t+1)  4^(t+2) = 128, find t [#permalink]
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15 Nov 2013, 02:11
Why can't you solve it this way?
4^{2t+1}  4^{t+2} = 128
(2^{2})^{2t+1}  (2^{2})^{t+2} = 2^{7}
2^{4t+2}  2^{2t+4} = 2^{7}
4t+2  (2t+4) = 7
4t+2  2t4 = 7
2t  2 = 7
2t = 9
t = 4.5
Is there something I'm missing here?



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Re: Given: 4^(2t+1)  4^(t+2) = 128, find t [#permalink]
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15 Nov 2013, 02:16



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Re: Given: 4^(2t+1)  4^(t+2) = 128, find t [#permalink]
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15 Nov 2013, 02:18
Bunuel wrote: Daddydekker wrote: Why can't you solve it this way?
4^{2t+1}  4^{t+2} = 128
(2^{2})^{2t+1}  (2^{2})^{t+2} = 2^{7}
2^{4t+2}  2^{2t+4} = 2^{7}
4t+2  (2t+4) = 7
4t+2  2t4 = 7
2t  2 = 7
2t = 9
t = 4.5
Is there something I'm missing here? \(2^x2^y=2^z\) does not mean that \(xy=z\). True, but bases are the same. So you can add right?



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Re: Given: 4^(2t+1)  4^(t+2) = 128, find t [#permalink]
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Updated on: 15 Nov 2013, 02:21
Daddydekker wrote: Bunuel wrote: Daddydekker wrote: Why can't you solve it this way?
4^{2t+1}  4^{t+2} = 128
(2^{2})^{2t+1}  (2^{2})^{t+2} = 2^{7}
2^{4t+2}  2^{2t+4} = 2^{7}
4t+2  (2t+4) = 7
4t+2  2t4 = 7
2t  2 = 7
2t = 9
t = 4.5
Is there something I'm missing here? \(2^x2^y=2^z\) does not mean that \(xy=z\). True, but bases are the same. So you can add right? Nah, to get x  y = z the equation would have to be: \(\frac{2^x}{2^y}\)= 2^z I initially solved it the same way you did as well and got 4.5.
Originally posted by NvrEvrGvUp on 15 Nov 2013, 02:19.
Last edited by NvrEvrGvUp on 15 Nov 2013, 02:21, edited 1 time in total.



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Re: Given: 4^(2t+1)  4^(t+2) = 128, find t [#permalink]
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15 Nov 2013, 02:20



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Re: Given: 4^(2t+1)  4^(t+2) = 128, find t [#permalink]
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15 Nov 2013, 02:23
Bunuel wrote: Daddydekker wrote: Bunuel wrote: \(2^x2^y=2^z\) does not mean that \(xy=z\).
True, but bases are the same. So you can add right? I don't understand what you mean. Please go through this topic for basics: mathnumbertheory88376.htmlHe's thinking about this: \(\frac{2^x}{2^y}\)= 2^z, which becomes 2^(xy) = 2^z In this case, we can say that x  y = z since there are only 2 exponents on two sides of the equal sign with the same base



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Re: Given: 4^(2t+1)  4^(t+2) = 128, find t [#permalink]
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17 Nov 2013, 14:07
This is a crazy problem even after one discovers the hidden quadratic.



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Re: Given: 4^(2t+1)  4^(t+2) = 128, find t [#permalink]
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26 Feb 2014, 02:25
4^(2t+1)  4^(t+2) = 128 Dividing by 4 to the equation 4^(2t)  4^t . 4 = 32 (4^t) ^ 2  4^t . 4  32 = 0 Let 4^t = a a^2  4a  32 = 0 Roots are a = 8 & 4 (neglect 4 as ve) So, a = 8 4^x = 8 = 4 ^ ( 1 + 1/2) [ 8 can be written as 4 x 2 = 4^1 + 4^(1/2) ]So, x = 3/2 = 1.5 = Answer
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Re: Given: 4^(2t+1)  4^(t+2) = 128, find t [#permalink]
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14 Aug 2015, 06:53
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NvrEvrGvUp wrote: Given: 4^(2t+1)  4^(t+2) = 128 Find: t 128 = 2^7 = 2^7*(21) =2^82^7 4t +2=8 t = 1.5



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Given: 4^(2t+1)  4^(t+2) = 128, find t [#permalink]
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20 Sep 2016, 14:22
Given: 4^(2t+1)  4^(t+2) = 128 Find: t
if 2^82^7=256128=128 then 4^44^3.5=128 t=1.5



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Re: Given: 4^(2t+1)  4^(t+2) = 128, find t [#permalink]
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21 Sep 2016, 00:10
Bunuel wrote: NvrEvrGvUp wrote: Given: 4^(2t+1)  4^(t+2) = 128 Find: t \(4^{2t+1}  4^{t+2} = 128\); \(4*4^{2t}  4^2*4^{t} = 128\); \(4*(4^t)^2 16*4^t  128=0\); \((4^t)^2 4*4^t  32=0\); Solve for 4^t: \(4^t=4\) (discard, because 4^t cannot be negative) or \(4^t=8\). \(4^t=8\) > \(2^{2t}=2^3\) > \(2t=3\) > \(t=1.5\). Hope it's clear. I managed to get the same equation, but never noticed the quardratic in there. Thanks for the approach.



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Re: Given: 4^(2t+1)  4^(t+2) = 128, find t [#permalink]
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26 Oct 2017, 15:38
NvrEvrGvUp wrote: Given: 4^(2t+1)  4^(t+2) = 128 Find: t 4^(2t+1)  4^(t+2) = 128 4^2t x 4  4^t x 4^2 = 128 4(4^t)^2  16(4^t)  128 = 0 If we let x = 4^t, we have: 4x^2  16x  128 = 0 x^2  4x  32 = 0 (x  8)(x + 4) = 0 x = 8 or x = 4 Since x = 4^t, we have: 4^t = 8 or 4^t = 4 Notice that 4 is positive, so 4^t also will be positive for any values of t, and thus 4^t can’t be 4. We are left with 4^t = 8: 2^(2t) = 2^3 2t = 3 t = 3/2
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Re: Given: 4^(2t+1)  4^(t+2) = 128, find t
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