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Given: 4^(2t+1) - 4^(t+2) = 128, find t [#permalink ]

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15 Nov 2013, 01:52
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Given: 4^(2t+1) - 4^(t+2) = 128

Find: t

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Bunuel on 15 Nov 2013, 01:53, 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, 02:00
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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 - 2t-4 = 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 - 2t-4 = 7 2t - 2 = 7 2t = 9 t = 4.5 Is there something I'm missing here?

\(2^x-2^y=2^z\) does not mean that \(x-y=z\).

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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15 Nov 2013, 02:19

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 - 2t-4 = 7 2t - 2 = 7 2t = 9 t = 4.5 Is there something I'm missing here?

\(2^x-2^y=2^z\) does not mean that \(x-y=z\).

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.

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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:

\(2^x-2^y=2^z\) does not mean that \(x-y=z\).

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:

math-number-theory-88376.html He's thinking about this: \(\frac{2^x}{2^y}\)= 2^z, which becomes 2^(x-y) = 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

NvrEvrGvUp wrote:

Given: 4^(2t+1) - 4^(t+2) = 128

Find: t

128 = 2^7 = 2^7*(2-1) =2^8-2^7

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 2^8-2^7=256-128=128 then 4^4-4^3.5=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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