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

Question

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

Bunuel · Accepted Answer

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.

JeffTargetTestPrep · Answer

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

NvrEvrGvUp · Answer

Hidden quadratic. Got it, thanks Bunuel.

PareshGmat · Answer

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)

So, x = 3/2 = 1.5 = Answer

mathivanan · Answer

128 = 2^7 = 2^7*(2-1) =2^8-2^7
4t +2=8
t = 1.5

gokul5544 · Answer

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.

Kinshook · Answer

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

ZIX · Answer

Here's a non-quadratic approach:

4^{2t+1} - 4^{t+2} = 128  (factor out 4, and express it in terms of  2^x ) 
  2^{4t} - 2^{2t+2} = 2^5

Now, difference b/w exponents with base 2 would result in the same base exponent when:

2^2 - 2^1 = 2^1  
  2^3 - 2^2 = 2^2  
  2^4 - 2^3 = 2^3  
 ... 
 So only  2^{x+1} - 2^x  would give us  2^x

You can also test with other powers of 2:  
  2^4 - 2^1 = 14  (won't work)

Now coming back to the question,

2^{4t} - 2^{2t+2} = 2^5

4t = 5 + 1  
  t = 1.5

ablatt4 · Answer

simplify instantly into

2^4t+2 - 2^2T+4 = 2^7

2^8-2^7= 2^7

Find the value that makes each term 8 and 7 respectively

D

rajdeep212003 · Answer

I got this hope it will be helpful

vallequalle · Answer

Just try out the options and you got it in one minute, no?

Usernamevisible · Answer

GMAT Takeaway

Whenever you see 4^2t and 4^t, use a substitution. x=4^t
for further solution check  Bunuel  solution

Ranjit.C · Answer

This can be done without quadratic equations and by using powers of 2 and 4

4^(2t+1) − 4^(t+2) = 128
 
4^4 = 256
4^3 = 64
4^3.5 = 4^(7/2) = 128

So, on comparing =>
2t+1 = 4 or t+2 = 3.5
=> t = 1.5 (Option D)

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

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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My Rewards