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How many distinct integral value of $n$ satisfies the equation $2^{2n} - 3 \cdot (2^{n+2}) + 32 = 0 $ ?

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2 Answers 2

up vote 7 down vote accepted

HINT: make the substitution $x=2^{n}$ This will reduce your equation to a quadratic.

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Thanks for the hint, So it boils to $X^2 -12x +32 = 0$ solving ${{x -> 4}, {x -> 8}}$ which gives n=2,3. Hence, 2 distinct integral value of n is possible. –  Quixotic Oct 28 '10 at 5:16
    
Yep! That's what I got as well :) –  WWright Oct 28 '10 at 5:17

You should be able to prove a fairly small limit on n. Think about the fact that 3<2^2. Then you can just try them all up to there.

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