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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] New post 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.
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Re: Given: 4^(2t+1) - 4^(t+2) = 128, find t [#permalink] New post 15 Nov 2013, 01:00
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Expert's post
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] New post 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] New post 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] New post 15 Nov 2013, 01:16
Expert's post
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.
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Re: Given: 4^(2t+1) - 4^(t+2) = 128, find t [#permalink] New post 15 Nov 2013, 01:18
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] New post 15 Nov 2013, 01:19
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.

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] New post 15 Nov 2013, 01:20
Expert's post
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?


I don't understand what you mean. Please go through this topic for basics: math-number-theory-88376.html
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Given: 4^(2t+1) - 4^(t+2) = 128, find t [#permalink] New post 15 Nov 2013, 01:23
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] New post 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] New post 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] 26 Feb 2014, 01:25
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