If n is a positive integer, is the value of b - a at least twice the value of 3^n - 2^n?
(1) a= 2^(n+1) and b= 3^(n+1)
(2) n = 3
(1) a= 2^(n+1) and b= 3^(n+1)
(2) n = 3
Thanks
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Airan
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The problem is highlighted above in blue.ShaneK wrote:Need a little help..
Is b-a greater than twice the value of 3^n - 2^n?
b - a > 2(3^n - 2^n)
b - a > 6^n - 4^n
Statement I:
b = 3^(n+1) a = 2^(n+1)
Added to Q Stem:
3^(n+1) - 2^(n+1) > 6^n - 4^n ?
If n = 1, 9 - 4 > 6 - 4 YES
If n = 2, 27 - 8 > 36 - 16 NO
.. I think I've been studying a bit too long today. What the heck am I missing?
Hey Shane,ShaneK wrote: Side note: If you double the coefficient of an exponent, are you essentially quadrupling the answer?
i.e. 3^2 = 9, 6^2 = 36.. this makes sense, because by doubling, you've added 2^n.. so you're not always quadrupling, you're 2^n power'ing, because 6^n is 2^n(3^n).
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