by tim Fri Dec 19, 2014 1:38 am
You can't just strip out the 2's in this case since you're adding two different powers on the right side of the equation. What you want to do is this:
128 = 4^(2t + 1) – 4^(t + 2)
pull out the constants:
128 = 4*(4^t)^2 - 16*4^t
divide everything by 4:
32 = (4^t)^2 - 4*4^t
now substitute x for 4^t:
32 = x^2 - 4x
x^2 - 4x - 32 = 0
(x-8)(x+4) = 0
x = 8 or -4
4^t = 8 or -4
but 4^t can never be negative, so 4^t = 8. at this point you can turn things into powers of 2:
4^t = 8
2^(2t) = 2^3
2t = 3
t = 3/2
the key on this one is to recognize that this is actually a quadratic. you can notice this because of the 4^t and 4^(2t). of course, you can also guess and check on this one; after you try 1 and 2 you should realize that the answer is between those numbers, and 1.5 is the next logical thing to try.
Tim Sanders
Manhattan GMAT Instructor
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