Given: 4^(2t+1) - 4^(t+2) = 128, find t : GMAT Problem Solving (PS)
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# Given: 4^(2t+1) - 4^(t+2) = 128, find t

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

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15 Nov 2013, 00:52
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Given: 4^(2t+1) - 4^(t+2) = 128
Find: t

[Reveal] Spoiler:
OA = 1.5

Last edited by Bunuel on 15 Nov 2013, 00:53, 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, 01:00
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NvrEvrGvUp wrote:
Given: 4^(2t+1) - 4^(t+2) = 128
Find: t

[Reveal] 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]

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15 Nov 2013, 01:16
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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 $$x-y=z$$.
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Re: Given: 4^(2t+1) - 4^(t+2) = 128, find t [#permalink]

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15 Nov 2013, 01:20
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Bunuel 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?

I don't understand what you mean. Please go through this topic for basics: math-number-theory-88376.html
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Re: Given: 4^(2t+1) - 4^(t+2) = 128, find t [#permalink]

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15 Nov 2013, 01:02
Bunuel wrote:
NvrEvrGvUp wrote:
Given: 4^(2t+1) - 4^(t+2) = 128
Find: t

[Reveal] 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]

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15 Nov 2013, 01:11
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, 01:18
Bunuel 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, 01:19
Bunuel 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.

Last edited by NvrEvrGvUp on 15 Nov 2013, 01:21, 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, 01:23
Bunuel 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, 13:07
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, 01:25
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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12 Aug 2015, 15:02
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Re: Given: 4^(2t+1) - 4^(t+2) = 128, find t [#permalink]

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14 Aug 2015, 05:53
NvrEvrGvUp wrote:
Given: 4^(2t+1) - 4^(t+2) = 128
Find: t

[Reveal] 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]

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20 Sep 2016, 11:15
Hello from the GMAT Club BumpBot!

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

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20 Sep 2016, 13:22
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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20 Sep 2016, 23:10
Bunuel wrote:
NvrEvrGvUp wrote:
Given: 4^(2t+1) - 4^(t+2) = 128
Find: t

[Reveal] 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.

I managed to get the same equation, but never noticed the quardratic in there. Thanks for the approach.
Re: Given: 4^(2t+1) - 4^(t+2) = 128, find t   [#permalink] 20 Sep 2016, 23:10
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# Given: 4^(2t+1) - 4^(t+2) = 128, find t

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