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Re: Solving exponents II [#permalink]
15 Nov 2010, 14:18
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This has always been a favorite problem of mine - the first time I saw it, an instructor had blanked on how to solve it and emailed me a photo from his phone asking for help. He had excused himself from a tutoring session and needed me to explain how to solve it (correctly, of course) within a few minutes, so the pressure was on!
Bunuel's explanation is perfect (as always), but when the pressure was on and I wasn't exactly thinking about factoring, I did this instead - I looked to see if there were a pattern in the subtraction at left (2 to an exponent minus 2 to another exponent, two less) that would always produce 3*something on the right. So I did:
x = 3 and x-2 = 1 2^3 - 2^1 = 8 - 2 = 6
And 6 = 3(2^1), so I had a start.
x = 4 and x-2 = 2 2^4 - 2^2 = 16 - 4 = 12
And 12 = 3(2^2), so the pattern held
x = 5 and x-2 = 3 2^5 - 2^3 = 32 - 8 = 24
And 24 = 3(2^3), and the pattern became clear... The operation at left was always producing 3*2^(x-2) as its answer, so if x-2 = 13, then x = 15.
Strategically, using small numbers to establish patterns works pretty well when huge numbers (like 3(2^13)) are in play, and when exponents are involved (exponents are essentially just repetitive multiplication, so there are bound to be some repetitive patterns involved). If you can factor like Bunuel did, that's a great way to go...but I'd recommend having the "prove patterns w/ small numers" ideology in your arsenal! _________________
Re: If 2^x - 2^(x-2) = 3*2^13, what is the value of x? [#permalink]
15 Jun 2014, 07:14
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