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

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NvrEvrGvUp
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Given: 4^(2t+1) - 4^(t+2) = 128, find t [#permalink] New post   Updated on: 22 Jul 2022, 02:44
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Given: 4^(2t+1) - 4^(t+2) = 128. Find: t

A. -1.5
B. -0.5
C. 0.5
D. 1.5
E. 2.5
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D

Originally posted by NvrEvrGvUp on 15 Nov 2013, 01:52.
Last edited by Bunuel on 22 Jul 2022, 02:44, edited 2 times in total.
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Re: Given: 4^(2t+1) - 4^(t+2) = 128, find t [#permalink] New post   15 Nov 2013, 02:00
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NvrEvrGvUp wrote:
Given: 4^(2t+1) - 4^(t+2) = 128
Find: t

Show Spoiler
OA = 1.5


\(4^{2t+1} - 4^{t+2} = 128\);

\(4*4^{2t} - 4^2*4^{t} = 128\);

\(4*(4^t)^2 -16*4^t - 128=0\);

\((4^t)^2 -4*4^t - 32=0\);

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.
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Re: Given: 4^(2t+1) - 4^(t+2) = 128, find t [#permalink] New post   26 Oct 2017, 15:38
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NvrEvrGvUp wrote:
Given: 4^(2t+1) - 4^(t+2) = 128
Find: t

Show Spoiler
OA = 1.5


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 [#permalink] New post   15 Nov 2013, 02:02
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Bunuel wrote:
NvrEvrGvUp wrote:
Given: 4^(2t+1) - 4^(t+2) = 128
Find: t

Show Spoiler
OA = 1.5


\(4^{2t+1} - 4^{t+2} = 128\);

\(4*4^{2t} - 4^2*4^{t} = 128\);

\(4*(4^t)^2 -16*4^t - 128=0\);

\((4^t)^2 -4*4^t - 32=0\);

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] New post   26 Feb 2014, 02:25
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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] New post   14 Aug 2015, 06:53
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NvrEvrGvUp wrote:
Given: 4^(2t+1) - 4^(t+2) = 128
Find: t

Show Spoiler
OA = 1.5

128 = 2^7 = 2^7*(2-1) =2^8-2^7
4t +2=8
t = 1.5
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Re: Given: 4^(2t+1) - 4^(t+2) = 128, find t [#permalink] New post   24 Jul 2022, 11:16
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Given: 4^(2t+1) - 4^(t+2) = 128. Find: t

Let 4^t = x

4x^2 - 16x - 128 = 0
x^2 - 4x - 32 = 0
(x-8)(x+4)=0
x=8 or x=-4
4^t = 8 ; t = 1.5

IMO D

Posted from my mobile device
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Re: Given: 4^(2t+1) - 4^(t+2) = 128, find t [#permalink] New post   24 Jul 2022, 09:52
I tired to visualize 128 as the difference of two powers of 4,

4^a - 4^b = 128

One way to make this split is

256 - 128 = 128

4^4- (4^0.5x4^3) = 128

2t+1= 4

t+2=3.5

Both give us 1.5.
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Re: Given: 4^(2t+1) - 4^(t+2) = 128, find t [#permalink] New post   09 Aug 2023, 10:41
Hello from the GMAT Club BumpBot!

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Re: Given: 4^(2t+1) - 4^(t+2) = 128, find t [#permalink]
09 Aug 2023, 10:41
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